Answers to Selected Exercises

Transcription

1 DalePhatANS_complete 8/18/04 10:30 AM Page 1049 Answers to Selected Exercises Chapter 1 Exam Preparation Exercises 1. a. v, b. i, c. viii, d. iii, e. iv, f. vii, g. vi, h. ii. 2. Analysis and specification, general solution (algorithm), verify. 3. Concrete solution (program), test. 4. The analysis and specification step within the problem-solving phase. 5. The steps never end. Changing the last step to say, Repeat from first step until graduation, converts the sequence into an algorithm that eventually ends, assuming the person will graduate from school some day. We can make sure that this loop ends by specifying a condition that will definitely occur. For example, Repeat from first step until last day of classes. 6. a. vi, b. ii, c. v, d. i, e. vii, f. iv. 7. The program can be compiled on different computer systems without modification. 8. The control unit directs the actions of the other components in the computer to execute program instruction in the proper order. 9. False. The editor is software rather than hardware. 10. False. Peripheral devices are external to the CPU and its main memory. 11. Yes. If the software license restricts use to a single computer or a single user, then this is a case of piracy, even though you split the cost. 12. a. ii, b. v, c. iii, d. vii, e. i, f. vi, g. iv. Chapter 1 Programming Warm-Up Exercises 1. Branches: Steps n1, n2, and n3. There are two loops: steps p, q, and r; and steps t and u. Steps a, c, and e are references to subalgorithms defined elsewhere.

2 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises 2. follow separate algorithm for filling glass with water and placing on counter If right-handed, pick up toothpaste tube in left hand and unscrew cap by turning counter-clockwise place toothpaste cap on counter transfer toothpaste tube to right hand and pick up toothbrush in left hand, holding it by its handle place open end of toothpaste tube against tips of brush bristles and squeeze toothpaste tube just enough to create a 0.5-inch-long extrusion of toothpaste on brush pull toothpaste tube away from brush and place tube on counter transfer toothbrush to right hand, holding it by its handle open mouth and insert brush, placing toothpaste-covered bristles against one tooth scrub tooth with brush by moving brush up and down 10 times reposition brush to an unscrubbed tooth repeat previous two steps until all teeth have been brushed spit toothpaste into sink put toothbrush in sink pick up glass in right hand fill mouth with water from glass swish water in mouth for five seconds spit water from mouth into sink repeat previous three steps three times Otherwise (if left-handed) pick up toothpaste tube in right hand and unscrew cap by turning counter-clockwise place toothpaste cap on counter transfer toothpaste tube to left hand and pick up toothbrush in right hand, holding it by its handle place open end of toothpaste tube against tips of brush bristles and squeeze toothpaste tube just enough to create a 0.5-inch-long extrusion of toothpaste on brush pull toothpaste tube away from brush and place tube on counter transfer toothbrush to left hand, holding it by its handle open mouth and insert brush, placing toothpaste-covered bristles against one tooth scrub tooth with brush by moving brush up and down 10 times reposition brush to an unscrubbed tooth repeat previous two steps until all teeth have been brushed spit toothpaste into sink put toothbrush in sink pick up glass in left hand fill mouth with water from glass swish water in mouth for five seconds spit water from mouth into sink repeat previous three steps three times Place glass on counter Follow separate algorithm for cleaning up after brushing teeth

3 DalePhatANS_complete 8/18/04 10:30 AM Page 1051 Answers to Selected Exercises The test for right- or left-handedness is a branch. Each branch contains two loops, one that iterates until all teeth are brushed and another that iterates rinsing four times. The first and last steps refer to separately defined subalgorithms. 4. Change step u to: u. Repeat step t nine times. Chapter 1 Case Study Follow-Up Exercises 1. a No b Yes c Yes d No 2. a Line = 6 b Line = 2 c Line = 7 d Line = 4 e Line = 4 3. Divide by 1000 and see if the remainder is I was not, but your answer may be different. 5. Keep adding 1 to the year and test it until you find the next leap year. Another algorithm would be to use the following: nextyear = year + (year % 4) divide the nextyear by 100 and if the remainder isn t 0, Write that nextyear is the next leap year Otherwise, divide the nextyear by 400 and if the remainder isn t 0 Write that (nextyear + 4) is the next leap year Otherwise nextyear is the next leap year 6. Line Number 1 if (year % 4!= 0) 2 return false; 3 else if (year % 100!= 0) 4 return true; 5 else if (year % 400!= 0) 6 return false; 7 else return true; Chapter 2 Exam Preparation Exercises 1. a. valid, b. invalid (hyphens not allowed), c. invalid (reserved words not allowed), d. valid, e. valid, f. valid, g. invalid (must begin with a letter), h. invalid (# is not allowed). 2. a. vi, b. ix, c. iv, d. v, e. vii, f. ii, g. i, h. x, i. iii, j. viii. 3. False.

4 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises 4. The template allows an identifier to begin with a digit. Identifiers can begin with a letter or an underscore, and digits may appear only in the second character position and beyond. 5. False. 6. True. 7. True. 8. Note that the second line concatenates two words without a separating space. Four score and seven years agoour fathers brought forth on this continent a new nation Bjarne Stroustrup named his new programming language C++ because it is a successor to the C programming language. 10. By preceding it with a backslash character (\ ). 11. If we forget the */ at the end of a comment, then any number of lines of program code can be inadvertently included in the comment. The // form avoids this by automatically terminating at the end of the line. 12. The // form of comment cannot span more than one line. It also cannot be inserted into the middle of a line of code because everything to the right of // becomes a comment. 13. << is the stream insertion operator, and is pronounced put to or is sent, as in cout is sent string No. The endl identifier is a manipulator, and is not a string value. 15. It tells the C++ preprocessor to insert the contents of a specified file at that point in the code. 16. std::cout << "Hello everybody!" << std::endl; 17. A block. 18. By splitting it into pieces that fit on a line, and joining them with the concatenation operator. 19. #include <iostream> #include <string> using namespace std; const string TITLE = "Dr."; int main() cout << "Hello " + TITLE + " Stroustrup!"; Chapter 2 Programming Warm-Up Exercises 1. Substitute the actual data in the following: cout << "July 4, 2004" << endl;

5 DalePhatANS_complete 8/18/04 10:30 AM Page 1053 Answers to Selected Exercises Note that the \" escape sequence is needed in six places here: cout << "He said, \"How is that possible?\"" << endl << "She replied, \"Using manipulators.\"" << endl << "\"Of course,\" he exclaimed!" << endl; 3. a. const string ANSWER = "True"; b. char middleinitial; c. string coursetitle; d. const char PERCENT = '%' 4. const string FIRST = "Your first name inserted here"; const string LAST = "Your last name inserted here"; // Insert your middle initial in place of A: const char MIDDLE = 'A'; 5. cout << PART1 << PART2 << PART1 << PART3; 6. PART1 + PART2 + PART1 + PART3 7. cout << "Yet the web of thought has no such creases" << endl; cout << "And is more like a weaver's masterpieces;" << endl; cout << "One step a thousand threads arise," << endl; cout << "Hither and thither shoots each shuttle," << endl; cout << "The threads flow on unseen and subtle," << endl; cout << "Each blow effects a thousand ties." << endl; 8. #include <iostream> #include <string> using namespace std; const string TITLE = "Rev."; const char FIRST = 'H'; const char MID 'G'; const char DOT = '.'; int main() cout << TITLE << FIRST << DOT << MID << DOT << " Jones"; 9. The solution to this problem is for the student to show that he or she has run the program correctly. It should output: *********************************** * * * Welcome to C++ Programming! * * * ***********************************

6 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises Chapter 2 Case Study Follow-Up 1. const string BLACK = "%%%%%%%%"; // Define a line of a black square 2. const string WHITE = "..."; // Define a line of a white square 3. Move the first five output statements to the end of the program. 4. const string BLACK = "**********"; // Define a line of a black square const string WHITE = " "; // Define a line of a white square 5. // Chessboard program // This program prints a chessboard pattern that is built up from // basic strings of white and black characters. #include <iostream> #include <string> using namespace std; const string BLACK = "********"; // Define a line of a black square const string WHITE = " "; // Define a line of a white square const string BLACK_BLANK = "** **"; // Define a constant of embedded // blanks int main () string whiterow; // A row beginning with a white square string blackrow; // A row beginning with a black square string blackblankrow; // A black first row with mixed squares string whiteblankrow; // A white first row with mixed squares // Create a white-black row by concatenating the basic strings whiterow = WHITE + BLACK + WHITE + BLACK + WHITE + BLACK + WHITE + BLACK; // Create a white-black row with interior blanks in the blacks whiteblankrow = WHITE + BLACK_BLANK + WHITE + BLACK_BLANK + WHITE + BLACK_BLANK + WHITE + BLACK_BLANK; // Create a black-white row with interior blanks in the blacks blackblankrow = BLACK_BLANK + WHITE + BLACK_BLANK + WHITE + BLACK_BLANK + WHITE + BLACK_BLANK + WHITE;

7 DalePhatANS_complete 8/18/04 10:30 AM Page 1055 Answers to Selected Exercises 1055 // Create a black-white row by concatenating the basic strings blackrow = BLACK + WHITE + BLACK + WHITE + BLACK + WHITE + BLACK + WHITE; // Print five white-black rows cout << whiterow << endl; cout << whiterow << endl; cout << whiteblankrow << endl; cout << whiteblankrow << endl; cout << whiterow << endl; // Print five black-white rows cout << blackrow << endl; cout << blackrow << endl; cout << blackblankrow << endl; cout << blackblankrow << endl; cout << blackrow << endl; // Print five white-black rows cout << whiterow << endl; cout << whiterow << endl; cout << whiteblankrow << endl; cout << whiteblankrow << endl; cout << whiterow << endl; // Print five black-white rows cout << blackrow << endl; cout << blackrow << endl; cout << blackblankrow << endl; cout << blackblankrow << endl; cout << blackrow << endl; // Print five white-black rows cout << whiterow << endl; cout << whiterow << endl; cout << whiteblankrow << endl; cout << whiteblankrow << endl; cout << whiterow << endl;

8 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises // Print five black-white rows cout << blackrow << endl; cout << blackrow << endl; cout << blackblankrow << endl; cout << blackblankrow << endl; cout << blackrow << endl; // Print five white-black rows cout << whiterow << endl; cout << whiterow << endl; cout << whiteblankrow << endl; cout << whiteblankrow << endl; cout << whiterow << endl; // Print five black-white rows cout << blackrow << endl; cout << blackrow << endl; cout << blackblankrow << endl; cout << blackblankrow << endl; cout << blackrow << endl; return 0; 6. There are 64 characters per line, except for the results of Exercise 4, which would have 80 characters per line. Chapter 3 Exam Preparation Exercises 1. They are simple data types. 2. char, short, int, long. 3. Integer overflow. 4. E signifies an exponent in scientific notation. The digits to the left of the E are multiplied by 10 raised to the power given by the digits to the right of the E.

9 DalePhatANS_complete 8/18/04 10:30 AM Page 1057 Answers to Selected Exercises integer / constant / floating variable a. const int TRACKS_ON_DISK = 17; integer constant b. float timeoftrack; floating variable c. const float MAX_TIME_ON_DISK = 74.0; floating constant d. short tracksleft; integer variable e. float timeleft; floating variable f. long samplesintrack; integer variable g. const double SAMPLE_RATE = ; floating constant 6. Integer division with an integer result, and floating division with a floating result. 7. a. 21, b. 21.6, c. 13.0, d. 18, e. 20, f. 22, g () unary - [ * / % ] [ + - ] 9. True. 10. a. vi, b. ii, c. vii, d. iv, e. i, f. v, g. iii. 11. The ++ operator. 12. You write the name of a data type, followed by an expression enclosed in parentheses. 13. main returns an int value False. When used alone they are equivalent, but used within a larger expression, they can give different results (this is explained more in Chapter 10). 16. string::size_type 17. a. 29 b. 1 c. "ought to start with logic" d. 0 e. "log" f. 1 g. string::npos 18. It tells the stream to print floating-point numbers without using scientific notation. Chapter 3 Programming Warm-Up Exercises 1. (hours * 60 + minutes) * 60 + seconds 2. a. days / 7 b. days % 7 3. dollars * quarters * 25 + dimes *10 + nickels * 5 + pennies 4. float(dollars * quarters * 25 + dimes *10 + nickels * 5 + pennies) / count = count + 3; 6. a. 3 * X + Y b. A * A + 2 * B + C c. ((A + B)/(C D)) * (X / Y) d. ((A * A + 2 * B + C)/D) / (X * Y)

10 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises e. sqrt(fabs(a B)) f. pow(x, -cos(y)) 7. string temp; temp = sentence; first = temp.find("and"); // Get first location // Remove first section of string temp = temp.substr(first + 3, temp.length() (first + 3)); // Find second location of "and" in shortened string second = temp.find("and"); // Remove next section of string temp = temp.substr(second + 3, temp.length() (second + 3)); // Adjust second to account for deleting first part of string second = second + first + 3; // Locate third "and" and adjust for prior deletions third = temp.find("and") + second + 3; 8. startofmiddle = name.find(' ') + 1; 9. cout << "$" << fixed << setprecision(2) << setw(8) << money; 10. cout << setprecision(5) << setw(15) << distance; 11. #include <climits> cout << "INT_MAX = " << INT_MAX << " INT_MIN = " << INT_MIN; 12. //********************************************** // Celsius program // This program outputs the Celsius temperature // corresponding to a given Fahrenheit temperature //********************************************** #include <iostream> using namespace std; int main() const float FAHRENHEIT = 72.0;

11 DalePhatANS_complete 8/18/04 10:30 AM Page 1059 Answers to Selected Exercises 1059 float celsius; celsius = 5/9 * (FAHRENHEIT 32); cout << fixed << "Celsius equivalent of Fahrenheit " << FAHRENHEIT << " is " << celsius << " degrees."; return 0; Chapter 3 Case Study Follow-Up 1. cout << fixed << setprecision(2) << "For a loan amount of " << LOAN_AMOUNT << " with an interest rate of " << setprecision(4) << YEARLY_INTEREST << " and a " << NUMBER_OF_YEARS << " year mortgage, " << endl; cout << "your monthly payments are $" << setprecision(2) << payment << "." << endl; 2. Change the NUMBER_OF_YEARS to NUMBER_OF_MONTHS with the value 84. Remove numberofpayments and substitute NUMBER_OF_MONTHS where numberofpayments occurs. Change... const float LOAN_AMOUNT = ; // Amount of the loan const float YEARLY_INTEREST = ; // Yearly interest rate const int NUMBER_OF_MONTHS = 84; // Total number of payments int main() // local variables float monthlyinterest; // Monthly interest rate float payment; // Monthly payment // Calculate values monthlyinterest = YEARLY_INTEREST / 12; payment = (LOAN_AMOUNT * pow(monthlyinterest + 1, NUMBER_OF_MONTHS) * monthlyinterest)/(pow(monthlyinterest + 1, NUMBER_OF_MONTHS) - 1 ); // Output results cout << fixed << setprecision(2) << "For a loan amount of " << LOAN_AMOUNT << " with an interest rate of " << setprecision(4) << YEARLY_INTEREST << " and a "

12 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises << NUMBER_OF_MONTHS << " month mortgage, " << endl; cout << "your monthly payments are $" << setprecision(2) << payment << "." << endl; const float LOAN_AMOUNT = ; // Amount of the loan const float YEARLY_INTEREST = 5.24; // Yearly interest rate const int NUMBER_OF_MONTHS = 84; // Total number of payments int main() // local variables float monthlyinterest; // Monthly interest rate float payment; // Monthly payment // Calculate values monthlyinterest = YEARLY_INTEREST * 0.01 / 12; payment = (LOAN_AMOUNT * pow(monthlyinterest + 1, NUMBER_OF_MONTHS) * monthlyinterest)/(pow(monthlyinterest + 1, NUMBER_OF_MONTHS) - 1 ); // Output results cout << fixed << setprecision(2) << "For a loan amount of " << LOAN_AMOUNT << " with an interest rate of " << YEARLY_INTEREST << " and a " << NUMBER_OF_MONTHS << " month mortgage, " << endl; cout << "your monthly payments are $" << payment << "." << endl; float oneplus; // Monthly interest plus 1 // Calculate values monthlyinterest = YEARLY_INTEREST * 0.01 / 12; oneplus = monthlyinterest + 1; payment = (LOAN_AMOUNT * pow(oneplus, NUMBER_OF_MONTHS) * monthlyinterest/(pow(oneplus, NUMBER_OF_MONTHS) - 1;...

13 DalePhatANS_complete 8/18/04 10:30 AM Page 1061 Answers to Selected Exercises 1061 If an expression is duplicated more than twice, it probably should be calculated and saved. If an expression is used only twice but is a complex expression, it probably should be calculated and saved. The duplicate expression in this program is simple and is used only twice; it probably isn t worth the extra storage. Chapter 4 Exam Preparation Exercises 1. False. The two statements input the values in reverse order with respect to the single statement s inputs. 2. a. The insertion operator cannot be used with cin. b. The extraction operator cannot be used with cout. c. The right operand of the extraction operator must be a variable. d. The arguments are in reverse order. e. The function name should not have a capital L, and the arguments are reversed. 3. a = 70, b = 80, c = 30, d = 40, e = 50, f = a = 70, b = 80, c = 30, d = 0, e = 20, f = a. string1 = "January", string2 = "25," b. The reading marker is on the space following the comma. 6. a. string1 = "January 25, 2005" b. The reading marker is at the start of the next line. 7. a. string1 = "January", int1 = 25, char1 = ',', string2 = "2005" b. The reading marker is on the newline character at the end of the line. 8. It returns \n. 9. The first argument is a maximum number of characters to skip over. The second argument specifies a character that the function searches for. If it finds the character before skipping the maximum number of characters, then it stops with the reading marker on the following character. 10. Some variation in the answer is acceptable, as long as it is clear that the prompt for the inexperienced user is much more detailed. a. cout << "Enter a date in the format of mm/dd/yyyy." << " For example, for October 18, 1989, enter 10/18/1989."; b. cout << "Enter date, format mm/dd/yyyy."; 11. Include the header file fstream. Declare the file stream. Prepare the file with open. Specify the name of the file stream in each I/O statement. 12. ifstream and ofstream. 13. The first statement declares infile to be an input file stream. The second statement opens infile, associating it with a file called datafile.dat by the operating system. 14. The correction is to convert the name into a C string. ifstream indata; string name; cout << "Enter the name of the file: "); cin >> name; infile.open(name.c_str());

14 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises 15. The input operation is ignored. 16. False. The file immediately enters the fail state, and I/O operations on it are simply ignored. 17. False. If the file doesn t exist, it can open in the fail state. 18. The answer can vary, but should at least include vehicle, customer, and agency. A little more thought might add garage, repair shop, contract, and so on. 19. The class. 20. a. A step for which the implementation details are fully specified. b. A step for which some implementation details remain to be specified. c. A self-contained series of steps that solves a given problem or subproblem. d. A module that performs the same operation as the abstract step it defines. e. A property of a module in which the concrete steps are all directed to solving just one problem, and any significant subproblems are written as abstract steps. 21. The function. 22. a. no, b. yes, c. yes, d. no, e. no. 23. True. Chapter 4 Programming Warm-Up Exercises 1. cin >> int1 >> int2 >> int3; 2. cout << "Enter a name in the format: first middle last:"; cin >> first >> middle >> last; 3. cout << "Enter a name in the format: first middle last:"; getline(cin, name); 4. float number1; float number2; float number3; float average; cout << "Enter the first number and press return: "; cin >> number1; cout << "Enter the second number and press return: "; cin >> number2; cout << "Enter the third number and press return: "; cin >> number3; average = (number1 + number2 + number3) / 3.0; cout << "The average is: " << average << endl; 5. float number1; float number2; float number3; float average; cout << "Enter the three numbers separated by spaces: "; cin >> number1 >> number2 >> number3;

15 DalePhatANS_complete 8/18/04 10:30 AM Page 1063 Answers to Selected Exercises 1063 average = (number1 + number2 + number3) / 3.0; cout << "The average is: " << average << endl; 6. The names of the variables may differ from those shown here. Instead of using the ignore function to skip a single character, reading into a dummy char variable may also be used. a. int int1; char char1; float float1; cin >> int1 >> char1 >> float1; b. string string1; string string2; int int1; int int2; cin >> string1 >> int1 >> string2 >> int2; c. int int1; int int2; float float1; cin >> int1; cin.ignore(1, ','); cin >> int2; cin.ignore(1, ','); cin >> float1; d. char char1; char char2; char char3; char char4; cin >> char1; cin.ignore(1, ' '); cin >> char2; cin.ignore(1, ' '); cin >> char3; cin.ignore(1, ' '); cin >> char4; cin.ignore(1, ' '); e. float float1; cin.ignore(1, '$'); cin >> float1;

16 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises 7. In this solution the comma is included as part of the city name. cin >> streetnum >> street1 >> street2 >> town >> state >> zip; 8. #include <fstream> fstream temps; temps.open("temperatures.dat"); 9. An alternative approach is to declare one temp variable that is reused for each of six input statements, and keep a running total of the input values. float temp1; float temp2; float temp3; float temp4; float temp5; float temp6; float average; temps >> temp1 >> temp2 >> temp3 >> temp4 >> temp15 >> temp6; average = (temp1 + temp2 + temp3 + temp4 + temp5 + temp6) / 6.0; cout << "The average temperature is: " << average; 10. #include <iostream> #include <fstream> using namespace std; fstream indata; fstream outdata; const float PI = ; float radius; float circumference; float area; int main() indata.open("indata.dat"); outdata.open("outdata.dat"); indata >> radius; circumference = radius * 2 * PI; area = radius * radius * PI; cout << "For the first circle, the circumference is " << circumference << " and the area is " << area << endl; outdata << radius << " " << circumference << " " << area << endl; indata >> radius;

17 DalePhatANS_complete 8/18/04 10:30 AM Page 1065 Answers to Selected Exercises 1065 circumference = radius * 2 * PI; area = radius * radius * PI; cout << "For the second circle, the circumference is " << circumference << " and the area is " << area << endl; outdata << radius << " " << circumference << " " << area << endl; 11. Changed lines are underlined. #include <iostream> #include <fstream> using namespace std; fstream indata; fstream outdata; const float PI = ; float radius; float circumference; float area; string filename; int main() indata.open("indata.dat"); cout << "Enter the name of the file to open: "; cin >> filename; outdata.open(filename.c_str()); // The remainder of the program is unchanged 12. infile >> int1 >> int2 >> int3; 13. cout << "Enter the name of the file to open: "; cin >> filename; userfile.open(filename.c_str()); 14. Top level: Write letter Mail letter

18 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises Level 2 Write letter Write return address and date Write company address Write salutation Write body of letter Write closing Sign letter Mail letter Address envelope Write return address Attach stamp to envelope Fold letter in thirds Insert letter in envelope Seal envelope Place envelope in mail box 15. Letter and envelope are the main objects. Other objects include: return address, company address, salutation, letter body, closing, signature, stamp, and mail box. Chapter 4 Case Study Follow-Up 1. The answer will vary with each student. 2. // Output information in required formats outdata << socialnum << endl; outdata << " " << firstname << ' ' << middlename << ' ' << lastname << ' ' << endl; outdata << " " << lastname << ", " << firstname << ' ' << middlename << ' ' << endl; outdata << " " << lastname << ", " << firstname << ' ' << initial << ' ' << endl; outdata << " " << firstname << ' ' << initial << ' ' << lastname; 3. #include <iostream> // Access cin and cout #include <fstream> // Access open... // Declare and open files ifstream indata; ofstream outdata; string filename; cout << "Enter the name of the input file: ";

19 DalePhatANS_complete 8/18/04 10:30 AM Page 1067 Answers to Selected Exercises cin >> filename; indata.open(filename.c_str()); outdata.open("name.out"); 4. #include <iostream> // Access cin and cout #include <fstream> // Access open... // Declare and open files ifstream indata; ofstream outdata; string filename; cout << "Enter the name of the input file: "; cin >> filename; indata.open(filename.c_str()); cout << "Enter the name of the output file: "; cin >> filename; outdata.open(filename.c_str()); // Mortgage Payment Calculator program // This program determines the monthly payments on a mortgage given // the loan amount, the yearly interest, and the number of years. #include <iostream> #include <cmath> #include <iomanip> using namespace std; int main() // Input variables float loanamount; float yearlyinterest; int numberofyears; // local variables float monthlyinterest; int numberofpayments; float payment;

20 DalePhatANS_complete 8/18/04 10:30 AM Page Answers to Selected Exercises // Prompt for and read input values cout << "Input the total loan amount (ex: )" << endl; cin >> loanamount; cout << "Input the interest rate (ex: 6.25)" << endl; cin >> yearlyinterest; cout << "Enter the number of years for the loan (ex: 15)" << endl; cin >> numberofyears; // Calculate values monthlyinterest = yearlyinterest * 0.01/ 12.0; numberofpayments = numberofyears * 12; payment = (loanamount * pow(1 + monthlyinterest, numberofpayments) * monthlyinterest)/ ( pow(1 + monthlyinterest, numberofpayments) - 1 ); // Output results cout << fixed << setprecision(2) << "For a loan amount of " << loanamount << " with an interest rate of " << yearlyinterest << " and a " << numberofyears << " year mortgage, " << endl; cout << fixed << setprecision(2) << "your monthly payments are $" << payment << "." << endl; return 0; Chapter 5 Exam Preparation Exercises 1. The order in which the computer executes statements in a program. 2. False. They are predefined constants. 3. False. It is >=. 4. Because, in the collating sequence of the character set, the uppercase letters all come before the lowercase letters. 5. a. true b. true c. true d. false e. false

21 DalePhatANS_complete 8/18/04 10:31 AM Page 1069 Answers to Selected Exercises 1069 f. true g. true 6. a. true b. true c. true d. true e. true f. false g. false 7. Because conditional or short-circuit evaluation prevents the second part of the expression from being evaluated when someint is True. 9. False. The parentheses are required. 10. It outputs The data doesn't make sense." 11. It outputs The word may be "The" 12. Nothing. 13. Because the test is written with = instead of ==, so the expression is always false. 14. It outputs Very good 15. Very goodexcellent 16. Add braces to enclose the If-Then statement. 17.!inData 18. There is no limit, although a human reader finds it difficult to follow too many levels. Chapter 5 Programming Warm-Up Exercises 1. moon == "blue" moon == "Blue" 2.!inFile1 &&!infile2 3. if (infile) cin >> somestring; 4. if (year1 < year2 && month1 < month2 && day1 < day2) cout << month1 << "/" << day1 << "/" year1 << " comes before " << month2 << "/' << day2 << "/" << year2; else cout << month1 << "/" << day1 << "/" year1 << " does not come before " << month2 << "/' << day2 << "/" << year2; 5. if (year1 < year2 && month1 < month2 && day1 < day2) cout << month1 << "/" << day1 << "/" year 1 << " comes before " << month2 << "/' << day2 << "/" << year2; else cout << month1 << "/" << day1 << "/" year1 << " does not come before "

22 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises << month2 << "/' << day2 << "/" << year2; month1 = month2; day1 = day2; year1 = year2; 6. bool1 bool2 &&!(bool1 && bool2) 7. if (score < 0 score > 100) cout << "Score is out of range"; 8. if (score < 0 score > 100) cout << "Score is out of range"; else scoretotal = scoretotal + score; scorecount++; 9. infile1 >> value1; infile2 >> value2; if (!infile1 &&!infile2) cout << "File error."; else if (!infile1!infile2) if (infile1) outfile << value1; else outfile << value2; else if (value1 < value2); outfile << value1; infile1 >> value1; else outfile << value2; infile2 >> value2; 10. if (score > 100) cout << "Duffer."; else if (score > 80) cout << "Weekend regular."; else if (score > 72) cout << "Competitive player."; else if (score > 68) cout << "Turn pro!"; else

23 DalePhatANS_complete 8/18/04 10:31 AM Page 1071 Answers to Selected Exercises 1071 cout << "Time to go on tour!"; 11. if (count1 <= count2 && count1 <= count3) cout << count1; if (count1 == count2) cout << count2; if (count1 == count3) cout << count3; else if (count2 <= count3) cout << count2; if (count2 == count3) cout << count3; else cout << count3; 12. The conditional tests are using = instead of ==. Both branches have the error, but the second one doesn t output the message because the result of the assignment expression is 0, which is treated as false by the branch. maximum = 75; minimum = 25; if (maximum == 100) cout << "Error in maximum: " << maximum << endl; if (minimum == 0) cout << "Error in minimum: " << minimum << endl; 13. if (area < 0) area = -area; root = sqrt(area); else root = sqrt(area); 14. Values of temp that should be tried are: 213, 212, 211, 33, 32, Combinations of values for month, day, and year that should be tested are as follows: Month Day Year Tests first condition of first branch not taken Tests second condition of first branch not taken Tests first branch taken, second branch taken Tests first branch taken, second not taken Tests first branch taken, second not taken, third taken

24 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Chapter 5 Case Study Follow-Up 1. You could plug the BMI cut-off value and a value for weight into the formula and solve it for the height; however, the body mass index is calculated using real arithmetic. The number of decimal places must be very accurate to get the exact BMI value for which you are looking. A better choice is to round the BMI value so that you are testing two integers. 2. There is no unique answer, but the following values will work. 120, , , , // BMI Program // This program calculates the body mass index (BMI) given a weight // in pounds and a height in inches and prints a health message // based on the BMI. Input in metric measures. #include <iostream> using namespace std; int main() float weight; // Weight in kilograms float height; // Height in meters float bodymassindex; // Appropriate BMI bool dataareok; // True if data non-negative // Prompt for and input weight and height cout << "Enter your weight in kilograms. " << endl; cin >> weight; cout << "Enter your height in meters. " << endl; cin >> height; // Test Data if (weight < 0 height < 0) dataareok = false; else dataareok = true;

25 DalePhatANS_complete 8/18/04 10:31 AM Page 1073 Answers to Selected Exercises 1073 if (dataareok) // Calculate body mass index bodymassindex = weight / (height * height); // Print message indicating status cout << "Your body mass index is " << bodymassindex << ". " << endl; cout << "Interpretation and instructions. " << endl; if (bodymassindex < 18.5) cout << "Underweight: Have a milkshake." << endl; else if (bodymassindex < 25.0) cout << "Normal: Have a glass of milk." << endl; else if (bodymassindex < 30.0) cout << "Overweight: Have a glass of iced tea." << endl; else cout << "Obese: See your doctor." << endl; else cout << "Invalid data; weight and height must be positive." << endl; return 0; 4. // BMI Program // This program calculates the body mass index (BMI) given a weight // in pounds and a height in feet and inches and prints a health // based message on the BMI. Input in English measures. #include <iostream> using namespace std; int main() const int BMI_CONSTANT = 703; float weight; float inches; float feet; float height; float bodymassindex; bool dataareok; // Constant in nonmetric formula // Weight in pounds // Remainder of height in inches // Height in feet // Complete height in inches // Appropriate BMI // True if data non-negative

26 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises // Prompt for and input weight and height cout << "Enter your weight in pounds: " ; cin >> weight; cout << "Enter your height in feet and inches. " << endl; cout << "Feet: " ; cin >> feet; cout << "Inches: " ; cin >> inches; height = feet * 12 + inches; // Test Data if (weight < 0 height < 0) dataareok = false; else dataareok = true; if (dataareok) // Calculate body mass index bodymassindex = weight * BMI_CONSTANT / (height * height); // Print message indicating status cout << "Your body mass index is " << bodymassindex << ". " << endl; cout << "Interpretation and instructions. " << endl; if (bodymassindex < 18.5) cout << "Underweight: Have a milkshake." << endl; else if (bodymassindex < 25.0) cout << "Normal: Have a glass of milk." << endl; else if (bodymassindex < 30.0) cout << "Overweight: Have a glass of iced tea." << endl; else cout << "Obese: See your doctor." << endl; else cout << "Invalid data; weight and height " << "must be positive." << endl; return 0;

27 DalePhatANS_complete 8/18/04 10:31 AM Page 1075 Answers to Selected Exercises The weight and height are checked to be sure they are not negative before the BMI is calculated in this program. There are several other things that could be done. For example, if the weight is less than the height, then there is an error. In fact, if the weight is less than one and a half times the height, there is a good chance that there is an error. This condition produces very low BMIs. What about weights that are six or seven times the height? These produce very high BMIs. Another approach would be to put a check on the BMI itself. If it is below 10 or greater than 60, there is a good chance of input errors. Chapter 6 Exam Preparation Exercises 1. False. 2. False. 3. True. 4. a. ii, b. ix, c. iv, d. viii, e. vi, f. i, g. v, h. vii, i. iii. 5. Twelve times Twelve times. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 7. Twelve times. $1.00 $2.00 $3.00 $4.00 $5.00 $6.00 $7.00 $8.00 $9.00 $10.00 $11.00 $12.00

28 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 8. The @@@@@@@@@@@ 9. The output is: The code fails to process the first value on the file. The corrected code is: sum = 0; indata >> number; while (indata) sum = sum + number; indata >> number; 11. You could choose anything that s not a name. A blank line would be one possibility. If the program must read the name in three parts, then a series of nonalphabetic characters, separated appropriately, would work. For example: The output is: The corrected code is: number = 0; while (number < 10)

29 DalePhatANS_complete 8/18/04 10:31 AM Page 1077 Answers to Selected Exercises 1077 cout << number * 2 1 << " "; number++; 13. False. 14. What is the condition that ends the loop? How should the condition be initialized? How should the condition be updated? What is the process being repeated? How should the process be initialized? How should the process by updated? What is the state of the program on exiting the loop? 15. The sum is not reinitialized at the start of each inner loop, so it simply keeps accumulating the input values. The count isn t updated, so the flow of control doesn t leave the inner loop until the end-of-file condition is encountered. Here is the corrected code. while (indata) count = 1; sum = 0; while (count <= 5 && indata) cin >> number; sum = sum + number; count++; cout << sum / count << endl; Chapter 6 Programming Warm-Up Exercises 1. count = -10; while (count <= 10) cout << count << endl; count++; 2. count = 1; sum = 0; while (sum <= 10000) sum = sum + count; count++; cout << count 1;

30 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 3. count = 0; sum = 0; indata >> score; while (indata && count < 20) sum = sum + score; count++; indata >> score; if (count == 0) cout << "No data on file."; else cout << "Average is " << sum / count << endl; 4. count = 0; getline(chapter6, line); while (chapter6) if (line.find("code segment") < string::npos) count++; getline(chapter6, line); cout << "The string was found " << count << " times." 5. cin >> response; while (!(response == "Yes" response == "yes" response == "No" response == "no")) cout << "Invalid response. Please enter \"Yes\" or \"No\"."; cin >> response; yes = response == "Yes" response == "yes"; 6. cout << "Sun Mon Tue Wed Thu Fri Sat" << endl; count = 1; while (count <= startday) //Print blanks in first week cout << " "; count++; daynumber = 1; while (daynumber <= days) //Print day numbers for month while (count <= 7) //Print for one week in month if (daynumber <= days)

31 DalePhatANS_complete 8/18/04 10:31 AM Page 1079 Answers to Selected Exercises 1079 cout << setw(3) << daynumber << " "; count++; daynumber++; cout << endl; count = 1; 7. a. startday = (startday + days) % 7; b. We would need the number of days in each successive month. 8. char onechar; int ecount; int charcount; ecount = 0; textdata >> onechar; charcount = 0; while (textdata) charcount++; if (onechar == 'z') ecount++; textdata >> onechar; cout << "Percentage of letter 'z': " << float(ecount) / charcount * 100; 9. char onechar; int ecount; int charcount; ecount = 0; textdata >> onechar; charcount = 0; while (textdata && charcount < 10000) charcount++; if (onechar == 'z') ecount++; textdata >> onechar; cout << "Percentage of letter 'z': " << float(ecount) / charcount * 100; 10. first = 1; second = 1; cout << first << endl << second << endl; while (second < 30000)

32 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises next = first + second; first = second; second = next; cout << second << endl; 11. first = 1; second = 1; cout << setw(5) << 1 << setw(8) << first << endl << setw(5) << 1 << setw(8) << second << endl; position = 2; while (second < 30000) next = first + second; first = second; second = next; position++; cout << setw(5) << position << setw(8) << second << endl; 12. cout << "Enter number of stars: "; cin >> number; count = 1; while (count <= number) cout << '*'; count++; ) cout << endl; 13. The loop ends when the specified number of stars has been printed. The condition is initialized by getting the number of stars, and setting the loop iteration counter to 1. The condition is updated by incrementing the iteration counter at the end of each iteration. The process being repeated is the printing of a single star. The process does not require any initialization or update. Upon exiting the loop, the specified number of stars has been printed, and the iteration counter is equal to the number plus one. 14. indata >> number; while (indata) count = 1; while (count <= number) cout << '*'; count++;

33 DalePhatANS_complete 8/18/04 10:31 AM Page 1081 Answers to Selected Exercises 1081 cout << endl; indata >> number; 15. The code should be tested with an empty file, a file that contains one value, and a file that contains multiple values. The values should include a negative integer, zero, and positive integers. Chapter 6 Case Study Follow-Up 1. if (current >= high) // Check for new high 2. if (preceding <= low) 3. Yes; a priming read is used because the first value read is special and needs to be processed outside the loop. Normally, a priming read is not used with a count-controlled loop. 4. Count-controlled. 5. Changing the input to come from the keyboard is easy: cout replaces all references to the file name in output statements and all other references to the file are removed. However, because the output is also going to the screen, it will be intermingled with the prompts and the input. If you are careful, you can make the screen readable, but if the input is from the keyboard, it is better for the output to go to a file. 6. Often the amount of data determines whether it should be entered into a file rather than entered at the keyboard. Another factor is where the data is coming from. In this case, values are being recorded by a machine, so it is logical to have them go directly to a file. 7. We should try data sets that present the highest value and lowest dip in different places and in different orders. For example, we should try one set with the highest reading as the first value and another set with it as the last. We should try test data with no dips (all the readings equal), with multiple dips, and with the lowest dip before, after, and between other dips. Finally, we should try the program on some typical sets of readings and check the output with results we ve determined by hand. Chapter 7 Exam Preparation Exercises 1. It uses the keyword void instead of int, the function is not called main, and it does not contain a return statement. 2. False. The parameter name is optional. 3. Control returns after the last statement has executed, and returns to the statement immediately following the call. 4. a. v, b. iii, c. vi, d. i, e. iv, f. viii, g. ii, h. vii. 5. name and salary are reference parameters. age and level are value parameters. 6. The call must have six arguments, unless default parameters are used (we do not cover their use in this book, but they are mentioned). 7. The value parameter stores the new value but the argument is unaffected. The reference parameter stores the new value directly into the argument. 8. The type must come before the name of each parameter instead of after it. The & should be appended to the type instead of to the parameter name. The prototype should end with a semicolon.

34 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 9. False. Arguments must appear in the same order as the corresponding parameters. 10. Hiding a module implementation in a separate block with a formally specified interface. 11. Flow out, or in and out. 12. It should not contain a return statement. 13. The result is not passed back to the caller it should be a reference parameter instead of a local variable. 14. The result is not passed back to the caller result should be a reference parameter instead of a value parameter. 15. The x and y parameters should be value parameters their data flow is inward only. In the case of the y parameter, the value of the argument will be set to zero when the function returns. 16. False. Like any other variable declaration, it can be accessed only by statements within the block that follows its declaration. 17. True. 18. That the file contains valid data (only integers), and that it has at least one value (the mean is undefined for an empty data set). Chapter 7 Programming Warm-Up Exercises 1. void Max (/* in */ int num1, /* in */ int num2, /* out */ int& greatest) 2. void Max (int, int, int&); 3. void Max (/* in */ int num1, /* in */ int num2, /* out */ int& greatest) if (num1 > num2) greatest = num1; else greatest = num2; 4. void GetLeast (/* inout */ ifstream& infile, /* out */ int& lowest) 5. void GetLeast (ifstream&, int&); 6. void GetLeast (/* inout */ ifstream& infile, /* out */ int& lowest) int number; infile >> number; // Get a number lowest = INT_MAX; // Initially use greatest int while (infile) if (number < lowest)

35 DalePhatANS_complete 8/18/04 10:31 AM Page 1083 Answers to Selected Exercises 1083 lowest = number; infile >> number; // Remember lower number // Get next number 7. // Precondition: // File infile has been opened AND // header file climits has been included // File infile is at end-of-file AND // if the file is empty, lowest contains INT_MAX; // else smallest number on file infile is in lowest 8. void Reverse ( /* in */ string original, /* out */ string& lanigiro ) 9. void Reverse ( string, string& ); 10. void Reverse ( /* in */ string original, /* out */ string& lanigiro ) int count; count = 1; lanigiro = ""; while (count <= original.length()) lanigiro = lanigiro + original.substr(original.length() - count, 1); count++; ) 11. // Precondition: // Original must have a value // lanigiro contains the reverse of the string in original 12. void LowerCount ( /* out */ int& count) char inchar; count = 0; cin >> inchar; while (cin && inchar!= '\n') if (islower(inchar)) count++; cin >> inchar;

36 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 13. // Precondition: // The program has included the header cctype // A line has been read from cin AND // the number of lowercase letters on the line is in count 14. void GetNonemptyLine ( /* inout */ ifstream& infile, /* out */ string& line) getline(infile, line); while (infile && line == "") getline(infile, line); 15. void SkipToEmptyLine ( /* inout */ ifstream& infile, /* out */ int& skipped) skipped = 0; getline(infile, line); while (infile && line!= "") skipped++; getline(infile, line); 16. void TimeAdd ( /* inout */ int& days, /* inout */ int& hours, /* inout */ int& minutes, /* in */ int adddays, /* in */ int addhours, /* in */ int addminutes) int extrahour; int extraday; minutes = (minutes + addminutes) % 60; extrahour = (minutes + addminutes) / 60; hours = (hours + addhours + extrahour) % 24; extraday = (hours + addhours + extrahour) / 24; days = days + adddays + extraday; 17. void TimeAdd ( /* inout */ int& days, /* inout */ int& hours, /* inout */ int& minutes, /* inout */ int& seconds,

37 DalePhatANS_complete 8/18/04 10:31 AM Page 1085 Answers to Selected Exercises 1085 /* in */ int adddays, /* in */ int addhours, /* in */ int addminutes, /* in */ int addseconds) int extraminute; int extrahour; int extraday; seconds = (seconds + addseconds) % 60; extraminute = (seconds + addseconds) / 60; minutes = (minutes + addminutes + extraminute) % 60; extrahour = (minutes + addminutes + extraminute) / 60; hours = (hours + addhours + extrahour) % 24; extraday = (hours + addhours + extrahour) / 24; days = days + adddays + extraday; 18. void SeasonPrint (/* in */ int month, /* in */ int day); if (month == 12 && day >= 21) cout << "Winter"; else if ((month == 9 and day >= 21) month > 9) cout << "Fall"; else if ((month == 6 and day >= 21) month > 6) cout << "Summer"; else if ((month == 3 and day >= 21) month > 3) cout << "Spring"; else cout << "Winter"; Chapter 7 Case Study Follow-Up Exercises 1. //***************************************************************** // Mortgage Payment Tables program // This program prints a table showing loan amount, interest rate, // length of loan, monthly payments, and total cost of a mortgage. #include <iostream> #include <cmath> #include <iomanip> #include <fstream> using namespace std;

38 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises // Function prototypes void PrintHeading(ofstream&); void GetLoanAmount(float&); void GetRestOfData(float&, int&); void DeterminePayment(float, int, float, float&); void PrintResults(ofstream&, float, int, float, float); int main() // Input variables float loanamount; float yearlyrate; int numberofyears; float payment; ofstream dataout; dataout.open("mortgage.out"); PrintHeading(dataOut); // Prompt for and read input values GetLoanAmount(loanAmount); while (loanamount > 0.0) GetRestOfData(yearlyRate, numberofyears); DeterminePayment(loanAmount, numberofyears, yearlyrate, payment); PrintResults(dataOut, loanamount, numberofyears, yearlyrate, payment); GetLoanAmount(loanAmount); dataout.close(); return 0; void PrintHeading(ofstream& dataout) dataout << fixed << setprecision(2) << setw(12) << "Loan Amount" << setw(12) << "No. Years" << setw(15) << "Interest Rate" << setw(12) << "Payment" << setw(12) << "Total Paid" << endl;

39 DalePhatANS_complete 8/18/04 10:31 AM Page 1087 Answers to Selected Exercises 1087 void GetLoanAmount(float& loanamount) cout << "Input the total loan amount; " << "a negative value stops processing." << endl; cin >> loanamount; void GetRestOfData(float& interestrate, int& numberofyears) cout << "Input the interest rate." << endl; cin >> interestrate; cout << "Enter the number of years for the loan" << endl; cin >> numberofyears; void DeterminePayment(float loanamount, int numberofyears, float yearlyrate, float& payment) // Precondition: // Loan amount // Term of loan // Interest rate // Monthly payment // local variables float monthlyrate; int numberofpayments; monthlyrate = yearlyrate / 1200; numberofpayments = numberofyears * 12; payment = (loanamount * pow(1 + monthlyrate, numberofpayments) * monthlyrate) / (pow(1 + monthlyrate, numberofpayments) - 1 ); void PrintResults(ofstream& dataout, float loanamount, int numberofyears, float yearlyrate, float payment)

40 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises dataout << fixed << setprecision(2) << setw(12) << loanamount << setw(12) << numberofyears << setw(12) << yearlyrate << setw(15) << payment << setw(12) << numberofyears*12*payment << endl; 2. There is no correct answer that is easier to read. Clearly, values that do not make sense cause problems. inf and nan are C++ special symbols that mean infinite and undefined. Loan amount, number of years, and interest rate should be greater than zero. 3. Error checking has three parts: what is an error, where do you check for it, and what do you do when it occurs. In this case, a nonpositive value for any of the inputs is an error, and the errors should be checked for immediately after the values are input. What should you do when nonpositive values are input? You have several choices. You can prompt the user to re-enter the bad value, enclosing the reading within a loop. Of course, if a correct value is never input, an infinite loop occurs. Another option is to write an error message and stop the program; for example, you can add a Boolean variable to each input function that returns true if the value(s) are OK and false otherwise. After each call, this variable can be checked before the program continues. Because there are so many possible solutions, we do not show one. Check over yours and be sure you have the three parts. 4. void PrintResults( /* inout */ ofstream& dataout, // Output file /* in */ float loanamount, // Loan amount /* in */ int numberofyears, // Term of loan /* in */ float yearlyrate, // Interest rate /* in */ float payment ) // Payment float amountpaid;

41 DalePhatANS_complete 8/18/04 10:31 AM Page 1089 Answers to Selected Exercises 1089 amountpaid = numberofyears*12*payment; dataout << fixed << setprecision(2) << setw(12) << loanamount << setw(12) << numberofyears << setw(12) << yearlyrate << setw(15) << payment << setw(12) << amountpaid << setw(12) << amountpaid - loanamount << endl; void PrintHeading( /* inout */ ofstream& dataout ) dataout << fixed << setprecision(2) << setw(12) << "Loan Amount" << setw(12) << "No. Years" << setw(15) << "Interest Rate" << setw(12) << "Payment" << setw(12) << "Total Paid" << setw(12) << "Interest Paid << endl; Chapter 8 Exam Preparation Exercises 1. True. 2. False. It is local to the function s body. 3. False. It cannot be referenced before it is declared, and it cannot be referenced within any block that declares an identifier with the same name. 4. True. 5. a. v, b. iv, c. i, d. vii, e. viii, f. vi, g. ii, h. iii. 6. The combination of param2 being a reference parameter and the erroneous use of an assignment expression instead of an equality test in the first If statement results in the argument to param2 being assigned the value in param1 every time the function is called. 7. The function GetYesOrNo refers to the global variable a, so the program always outputs the response to the function s input request as the name. 8. The namespace is in global scope. 9. The std namespace is local to main and not accessible to the other functions. 10. a. Static, b. Automatic, c. Static. 11. int pi = ; 12. It is initialized once, when control first reaches the statement. 13. It is initialized each time control reaches it. 14. False. 15. If a is less than or equal to b, the function fails to execute a return statement. 16. The return expression is of type float, and the return type of the function is int. 17. A void function should not have a return statement with an expression. 18. It makes a side effect error more likely.

42 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Chapter 8 1. #include <iostream> Programming Warm-Up Exercises using namespace std; int Power (int, int); int main () int pow; int x; cout << "Enter power: "; cin >> pow; cout << "Enter value to be raised to power: "; cin >> x; cout << Power(x, pow); int Power(int x, int pow) int result = 1; while (pow > 0) result = result * x; pow--; return result; 2. bool Equals (/* in */ float x, /* in */ float y) 3. bool Equals (float, float); 4. bool Equals (/* in */ float x, /* in */ float y) return abs(x y) < ; 5. float ConeVolume (/* in */ float radius, /* in */ float height); 6. float ConeVolume (/* in */ float radius, /* in */ float height) return 1.0/3.0 * * radius * radius * height;

43 DalePhatANS_complete 8/18/04 10:31 AM Page 1091 Answers to Selected Exercises int GetLeast (/* inout */ ifstream& infile) int lowest; int number; infile >> number; // Get a number lowest = INT_MAX; // Initially use greatest int while (infile) if (number < lowest) lowest = number; // Remember lower number infile >> number; // Get next number return lowest; 8. string Reverse ( /* in */ string original) string lanigiro; int count; count = 1; lanigiro = ""; while (count <= original.length()) lanigiro = lanigiro + original.substr(original.length() - count, 1); count++; return lanigiro; ) 9. int LowerCount () char inchar; int count; count = 0; cin >> inchar; while (cin && inchar!= '\n') if (islower(inchar)) count++; cin >> inchar; return count;

44 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 10. float SquareKm (/* in */ float length, /* in */ float width); return length * 1.6 * width * 1.6; 11. bool Exahausted (/* in */ int filmrolls) static int total = 0; total = total + filmrolls; if (total > 1000) total = 0; return true; else return false; 12. string MonthAbbrev (/* in */ int month) if (month == 12) return "Dec"; else if (month == 11) return "Nov"; else if (month == 10) return "Oct"; else if (month == 9) return "Sep"; else if (month == 8) return "Aug"; else if (month == 7) return "Jul"; else if (month == 6) return "Jun"; else if (month == 5) return "May"; else if (month == 4) return "Apr"; else if (month == 3) return "Mar"; else if (month == 2) return "Feb"; else return "Jan"; 13. string MonthAbbrev (/* in */ int month) if (month == 12) return "Dec"; else if (month == 11) return "Nov"; else if (month == 10) return "Oct"; else if (month == 9) return "Sep"; else if (month == 8) return "Aug"; else if (month == 7) return "Jul"; else if (month == 6) return "Jun"; else if (month == 5) return "May"; else if (month == 4) return "Apr";

45 DalePhatANS_complete 8/18/04 10:31 AM Page 1093 Answers to Selected Exercises 1093 else if (month == 3) return "Mar"; else if (month == 2) return "Feb"; else if (month == 1) return "Jan"; else return "Inv"; 14. float RunningAvg (/* in */ float value) static float total = 0.0; static int count = 0; total = total + value; count++; return total/float(count); Chaper 8 1. Case Study Follow-Up Exercises Evaluate BloodPressure(Inout: healthprofile; In: name) Prompt for name s input Get data Evaluate input according to charts and print message Evaluate Input and Print Level 2 if (systolic < 120) Print on healthprofile Systolic reading is optimal else if (systolic < 130) Print on healthprofile Systolic reading is normal else if (systolic < 140) Print on healthprofile Systolic reading is high normal else if (systolic < 160) Print on healthprofile Systolic indicates hypertension Stage 1 else if (systolic < 180) Print on healthprofile Systolic indicates hypertension Stage 2 else Print on healthprofile Systolic indicates hypertension Stage 3 if (diastolic < 80) Print on healthprofile Diastolic reading is optimal else if (diastolic < 85) Print on healthprofile Diastolic reading is normal else if (diastolic < 90) Print on healthprofile Diastolic reading is high normal

46 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises else if (diastolic < 100) Print on healthprofile Diastolic indicates hypertension Stage 1 else if (diastolic < 110) Print on healthprofile Diastolic indicates hypertension Stage 2 else Print on healthprofile Diastolic indicates hypertension Stage 3 2. // EvaluateBloodPressure Driver program // This program provides an environment for testing the // EvaluateBloodPressure function in isolation from the Profile // program #include <iostream> #include <fstream> using namespace std; void EvaluateBloodPressure(ofstream&, string); int main() ofstream healthprofile; healthprofile.open("profilebp"); string name = "John J. Smith"; for (int test = 1; test <= 12; test++) EvaluateBloodPressure(healthProfile, name); healthprofile.close(); return 0; void EvaluateBloodPressure ( /* inout */ ofstream& healthprofile, // Output file /* in */ string name ) // Patient's name // Declare the health statistics int systolic; int diastolic; // Enter the patient's health statistics cout << "Enter the systolic blood pressure reading for" << name << ": "; cin >> systolic; cout << "Enter the diastolic blood pressure reading for " << name<< ": ";

47 DalePhatANS_complete 8/18/04 10:31 AM Page 1095 Answers to Selected Exercises 1095 cin >> diastolic; healthprofile << "Blood Pressure Profile " << endl; if (systolic < 120) healthprofile << " Systolic reading is optimal" << endl; else if (systolic < 130) healthprofile << " Systolic reading is normal" << endl; else if (systolic < 140) healthprofile << " Systolic reading is high normal" << endl; else if (systolic < 160) healthprofile << " Systolic indicates hypertension Stage 1" << endl; else if (systolic < 180) healthprofile << " Systolic indicates hypertension Stage 2" << endl; else healthprofile << " Systolic indicates hypertension Stage 3" << endl; if (diastolic < 80) healthprofile << " Diastolic reading is optimal" << endl; else if (diastolic < 85) healthprofile << " Diastolic reading is normal" << endl; else if (diastolic < 90) healthprofile << " Diastolic reading is high normal" << endl; else if (diastolic < 100) healthprofile << " Diastolic indicates hypertension Stage 1" << endl; else if (diastolic < 110) healthprofile << " Diastolic indicates hypertension Stage 2" << endl; else healthprofile << " Diastolic indicates hypertension Stage 3" << endl; 3. // Profile Program // This program inputs a name, weight, height, blood pressure // readings, and cholesterol values. If an input value is // negative, an error message is written. Appropriate health // messages are written for each of the input values.

48 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises #include <fstream> #include <iostream> #include <string> using namespace std; string Name(); void EvaluateCholesterol(ofstream& healthprofile, string name, bool& dataok); void EvaluateBMI(ofstream& healthprofile, string name, bool& dataok); void EvaluateBloodPressure(ofstream& healthprofile, string name, bool& dataok); int main() // Declare and open the output file ofstream healthprofile; healthprofile.open("profile"); string name; bool dataok; name = Name(); // Evaluate the patient's statistics healthprofile << "Patient's name " << name << endl; EvaluateCholesterol(healthProfile, name, dataok); EvaluateBMI(healthProfile, name, dataok); EvaluateBloodPressure(healthProfile, name, dataok); healthprofile << endl; healthprofile.close(); return 0; // Name function // This function inputs a name and returns it in first, // middle initial, and last order. string Name() // Declare the patient's name string firstname; string lastname; char middleinitial;

49 DalePhatANS_complete 8/18/04 10:31 AM Page 1097 Answers to Selected Exercises 1097 // Enter the patient's name cout << "Enter the patient's first name: "; cin >> firstname; cout << "Enter the patient's last name: "; cin >> lastname; cout << "Enter the patient's middle initial: "; cin >> middleinitial; return firstname + ' ' + middleinitial + ". " + lastname; // Cholesterol function // This function inputs HDL (good cholesterol) and LDL (bad // cholesterol) and prints out a health message based on their // values on file healthprofile. void EvaluateCholesterol ( /* inout */ ofstream& healthprofile, // Output file /* in */ string name, // Patient's name /* out */ bool& dataok) int HDL; int LDL; cout << "Enter HDL for " << name << ": "; cin >> HDL; cout << "Enter LDL for " << name << ": "; cin >> LDL; dataok = HDL > 0 && HDL > 0; if (!dataok ) cout << " Cholesterol data must be positive; test aborted. "; else float ratio = LDL/HDL; // Calculate ratio of LDL to HDD healthprofile << "Cholesterol Profile " << endl; // Print message based on HDL value if (HDL < 40) healthprofile << " HDL is too low" << endl; else if (HDL < 60) healthprofile << " HDL is okay" << endl;

50 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises else healthprofile << " HDL is excellent" << endl; // Print message based on LDL value if (LDL < 100) healthprofile << " LDL is optimal" << endl; else if (LDL < 130) healthprofile << " LDL is near optimal" << endl; else if (LDL < 160) healthprofile << " LDL is borderline high" << endl; else if (LDL < 190) healthprofile << " LDL is high" << endl; else healthprofile << " LDL is very high" << endl; if (ratio < 3.22) healthprofile << " Ratio of LDL to HDL is good" << endl; else healthprofile << " Ratio of LDL to HDL is not good" << endl; // BMI Function // This function inputs weight in pounds and height in inches and // calculates the body mass index (BMI prints a health message // based on the BMI. Input in English weights. void EvaluateBMI ( /* inout */ ofstream& healthprofile, // Output file /* in */ string name, // Patient's name /* out */ bool& dataok) const int BMI_CONSTANT = 703; float pounds; float inches; // Constant in English formula cout << "Enter the weight in pounds for " << name << ": "; cin >> pounds; cout << "Enter the height in inches for " << name << ": "; cin >> inches;

51 DalePhatANS_complete 8/18/04 10:31 AM Page 1099 Answers to Selected Exercises 1099 dataok = pounds > 0 && inches > 0; if (!dataok) cout << " BMI data must be positive; test aborted. "; else float bodymassindex = pounds * BMI_CONSTANT / (inches * inches); healthprofile << "Body Mass Index Profile" << endl; // Print health message based on bodymassindex healthprofile << " Body mass index is " << bodymassindex << ". " << endl; healthprofile << " Interpretation of BMI " << endl; if (bodymassindex < 18.5) healthprofile << " Underweight: BMI is too low" << endl; else if (bodymassindex < 24.9) healthprofile << " Normal: BMI is average" << endl; else if (bodymassindex < 29.9) healthprofile << " Overweight: BMI is too high" << endl; else healthprofile << " Obese: BMI is dangerously high" << endl; // Blood Pressure function // This function gets blood pressure readings (systolic/diastolic) // and prints out a health message based on their values // in file healthprofile. void EvaluateBloodPressure ( /* inout */ ofstream& healthprofile, // Output file /* in */ string name, // Patient's name /* out */ bool& dataok) // Declare the health statistics int systolic;

52 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises int diastolic; // Enter the patient's health statistics cout << "Enter the systolic blood pressure reading for " << name << ": "; cin >> systolic; cout << "Enter the diastolic blood pressure reading for " << name << ": "; cin >> diastolic; dataok = systolic > 0 && diastolic > 0; if (!dataok ) cout << " Blood pressure data must be positive;" << " test aborted. "; else healthprofile << "Blood Pressure Profile " << endl; if (systolic < 120) healthprofile << " Systolic reading is optimal" << endl; else if (systolic < 130) healthprofile << " Systolic reading is normal" << endl; else if (systolic < 140) healthprofile << " Systolic reading is high normal" << endl; else if (systolic < 160) healthprofile << " Systolic indicates hypertension Stage 1" << endl; else if (systolic < 180) healthprofile << " Systolic indicates hypertension Stage 2" << endl; else healthprofile << " Systolic indicates hypertension Stage 3" << endl; if (diastolic < 80) healthprofile << " Diastolic reading is optimal" << endl; else if (diastolic < 85) healthprofile << " Diastolic reading is normal" << endl;

53 DalePhatANS_complete 8/18/04 10:31 AM Page 1101 Answers to Selected Exercises 1101 else if (diastolic < 90) healthprofile << " Diastolic reading is high normal" << endl; else if (diastolic < 100) healthprofile << " Diastolic indicates hypertension Stage 1" << endl; else if (diastolic < 110) healthprofile << " Diastolic indicates hypertension Stage 2" << endl; else healthprofile << " Diastolic indicates hypertension Stage 3" << endl; 4. Only the main function changes. int main() // Declare and open the output file ofstream healthprofile; healthprofile.open("profile"); string name; bool dataok; name = Name(); // Evaluate the patient's statistics healthprofile << "Patient's name " << name << endl; EvaluateCholesterol(healthProfile, name, dataok); if (dataok) EvaluateBMI(healthProfile, name, dataok); if (dataok) EvaluateBloodPressure(healthProfile, name, dataok); if (!dataok) cout << "Bad data entered; program aborted." << endl; healthprofile << endl; healthprofile.close(); return 0;

54 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Chapter 9 Exam Preparation Exercises 1. True. 2. False. It has local scope. 3. False. Converting a While that may execute zero times to a Do-While requires surrounding the Do-While with an If statement. 4. False. Both Break and Continue are allowed in any of the looping statements. 5. While and For are pretest loops, and Do-While is a posttest loop. 6. Control jumps to the case matching the switch expression, but then proceeds to execute the remaining statements in all of the succeeding cases. 7. while (true) times. 9. A Do-While loop would be most appropriate. 10. A While statement would be the best choice. 11. OakMapleError 12. August outcount = -1; while (outcount < 2) incount = 3; while (incount > 0) cout << outcount + incount << endl; incount --; outcount++;

55 DalePhatANS_complete 8/18/04 10:31 AM Page 1103 Answers to Selected Exercises How now brown cow? 17. cout << "How now brown cow?"; Chapter 9 Programming Warm-Up Exercises 1. switch (day) case 0 : cout << "Sunday"; break; case 1 : cout << "Monday"; break; case 2 : cout << "Tuesday"; break; case 3 : cout << "Wednesday"; break; case 4 : cout << "Thursday"; break; case 5 : cout << "Friday"; break; case 6 : cout << "Saturday"; break; 2. switch (day) case 0 : cout << "Sunday"; break; case 1 : cout << "Monday"; break; case 2 : cout << "Tuesday"; break; case 3 : cout << "Wednesday"; break; case 4 : cout << "Thursday"; break; case 5 : cout << "Friday"; break; case 6 : cout << "Saturday"; break; default: cout << "Error"; 3. Note that Break statements aren t needed because each branch of the Switch exits the function, accomplishing the required effect. string DayOfWeek (/* in */ int day) switch (day) case 0 : return "Sunday"; case 1 : return "Monday"; case 2 : return "Tuesday"; case 3 : return "Wednesday"; case 4 : return "Thursday"; case 5 : return "Friday"; case 6 : return "Saturday"; default: return "Error";

56 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 4. for (int day = 6, day >= 0, day--) cout << DayOfWeek(day) << endl; 5. for (int day = 3, day <= 9, day++) cout << DayOfWeek(day % 7) << endl; 6. char response; do cout << "Please enter 'Y' or 'N': "; cin >> response; if (response!= 'Y' && response!= 'N') cout << "Invalid response. "; while (response!= 'Y' && response!= 'N'); 7. int SumInt (/* in */ int goal) // Assumes goal is positive int num = 0; int sum = 0; do num++; sum = sum + num; while (sum < goal); return num; 8. for (int row = 1; row <= 10; row++) for (int col = 1; col <= 10; col++) cout << setw(5) << row * col; cout << endl; 9. for (int line = 1; line <= 10; line++) for (int star = 1; star <= line; star++) cout << '*'; cout << endl; 10. int line; int star; line = 1; do

57 DalePhatANS_complete 8/18/04 10:31 AM Page 1105 Answers to Selected Exercises 1105 star = 1; do cout << '*'; star++; while (star <= line); cout << endl; line++; while (line <= 10); 11. int line; int star; line = 1; while (line <= 10) star = 1; while (star <= line) cout << '*'; star++; cout << endl; line++; 12. void Rectangle (/* in */ int height, /* in */ int width) for (int col = 1; col <= width; col++) cout << '*'; cout << endl; for (int row = 2; row <= height 1; row++) cout << '*'; for (int col = 2; col < width 1; col++) cout << ' '; cout << '*' << endl; for (int col = 1; col <= width; col++) cout << '*'; cout << endl;

58 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 13. int starthour = 3; int endhour = 7; int startmin = 15; int endmin = 30; int beginmin; int lastmin; for (int hour = starthour; hour <= endhour; hour++) if (hour == starthour) beginmin = startmin; else beginmin = 0; if (hour == endhour) lastmin = endmin; else lastmin = 59; for (int min = beginmin; min <= lastmin; min++) cout << hour << ':' << min << endl; 14. int starthour = 3; int endhour = 7; int startmin = 15; int endmin = 30; int beginmin; int lastmin; for (int hour = starthour; hour <= endhour; hour++) if (hour == starthour) beginmin = startmin; else beginmin = 0; if (hour == endhour) lastmin = endmin; else lastmin = 59; for (int min = beginmin; min <= lastmin; min++) for (int sec = 0; sec <= 59; sec++) cout << hour << ':' << min << ':' << sec << endl; 15. Exercise 13 outputs 255 lines, and Exercise 14 outputs 15,300 lines.

59 DalePhatANS_complete 8/18/04 10:31 AM Page 1107 Answers to Selected Exercises 1107 Chapter 9 Case Study Follow-Up Exercises 1. The first two clauses in the If statement could be combined with if ((isupper(character) (islower(character)) or if ((isalpha(character)) 2. Words are ended by punctuation marks, blanks, or end of lines. A sum of these gives an estimate of the number of words. 3. Average word length would be the total number of alphanumeric characters divided by the number of words. 4. Average sentence length, percentage of two-character words, percentage of three-character words, percentage of words over a certain threshold all of these would be good candidates. Chapter 10 Exam Preparation Exercises 1. False. bool cannot be unsigned. 2. True. 3. True. sizeof(int) returns 4 for a 32-bit int, and 8 for a 64-bit int. 4. True. 5. False. This is purely a matter of style. 6. char, short, int, long, bool. 7. A leading zero makes the literal a base 8 (octal) number. A leading 0X makes the literal a base 16 (hexadecimal) number. 8. The assignment operators. 9. With ++count, the result of the expression is the same as the new (incremented) value of count. With count++, the result of the expression is the old value of count (prior to being incremented). 10. When the type name is not a single word, as in unsigned int. 11. toupper or tolower. 12. 'g' Absolute error is a fixed value of allowable error, whereas relative error multiplies the allowable error by one of the values being compared. 15. The identifier VIOLET is declared twice in the same scope. 16. The value of the expression is Because the result of incrementing choice is an int, which cannot be assigned to an enum type. 18. Because we must use the same type name for arguments, parameters, and function prototypes. If the type doesn t have a name, we can t use it in this context. Chapter a. 5 b. 5L c. 5UL d. 05 e. 0X5 Programming Warm-Up Exercises

60 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 2. a F b c L d E0 3. a. days += 7; b. radius *= ; c. workhours -= 40; d. average /= count; 4. sizeof(long) * 8 5. a. a * b + c * d b. a * b / (c * d) c. a + b + c / (d + e) * f d. (a + b) / (c + d) * (e + f) e. -a + b <= c * d && a + b >= c d 6. cin >> charin; if (isdigit( charin)) numin = charin '0'; 7. int ecount; string::size_type pos; ecount = 0; for (pos = 0; pos < inline.length(); pos++) if (inline[pos] == 'e') ecount++; 8. int MonthNum (/* in */ string month) switch (tolower(month[0])) case 'a': switch (tolower(month[1])) case 'p': return 4; case 'u': return 8; default : return 0; case 'd': return 12; case 'f': return 2; case 'j': switch (tolower(month[3])) case 'e': return 6; case 'u': return 1; case 'y': return 7; default : return 0;

61 DalePhatANS_complete 8/18/04 10:31 AM Page 1109 Answers to Selected Exercises 1109 case 'm': switch (tolower(month[2])) case 'r': return 3; case 'y': return 5; default : return 0; case 'n': return 11; case 'o': return 10; case 's': return 9; default : return 0; 9. if (tolower(ampm[0]) == 'p') hours = hours + 12; 10. abs(balance audit) <= abs(year epoch) <= epoch * enum Planets MERCURY, VENUS, EARTH, MARS, JUPITER, SATURN, URANUS, NEPTUNE, PLUTO; 13. Planets toplanet (/* in */ string name) switch (tolower(name[0])) case 'm': switch (tolower(name[1])) case 'e': return MERCURY; case 'a': return MARS; default : return EARTH; case 'v': return VENUS; case 'e': return EARTH; case 'j': return JUPITER; case 's': return SATURN; case 'u': return URANUS; case 'n': return NEPTUNE; case 'p': return PLUTO; default : return EARTH; 14. string PlanetToString (/* in */ Planet sphere) switch (sphere) case MERCURY: return "Mercury"; case VENUS: return "Venus";

62 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises case EARTH: return "Earth"; case MARS: return "Mars"; case JUPITER: return "Jupiter"; case SATURN: return "Saturn"; case URANUS: return "Uranus"; case NEPTUNE: return "Neptune"; case PLUTO: return "Pluto"; default : return "Error"; 15. for (Planet sphere = MERCURY; sphere <= PLUTO; sphere = Planet(sphere + 1)) cout << PlanetToString(sphere) << endl; Chapter 10 Case Study Follow-Up Exercises 1. The counters could be defined globally before function main, where the entire program could access them. However, this is considered bad style. In the next chapter, we cover the record structure, which is a way that these variables could be bound into one structure and passed as a single variable. 2. If a hyphen is immediately followed by an end-of-line symbol, then it is used to divide a word. Of course there might be an occasion when a hyphen just happens to fall on the end of a line. Without access to a dictionary, you can never be sure. 3. // Style Program // A stylistic analysis of the following features of text is // computed: // number of words // average word length // number of sentences // average sentence length // number of uppercase letters // number of lowercase letters // number of digits #include <fstream> #include <iostream> #include <iomanip> #include <cctype> using namespace std; enum Features UPPER, LOWER, DIGIT, IGNORE, EOW, EOS, COMPOUND;

63 DalePhatANS_complete 8/18/04 10:31 AM Page 1111 Answers to Selected Exercises 1111 void OpenFiles(ifstream&, ofstream&); Features Decode(char character); void ProcessCharacter(char, int&, int&, int&, int&, int&, int&); void PrintTable(ofstream& table, int, int, int, int, int, int); int main() // Prepare files for reading and writing ifstream text; ofstream table; OpenFiles(text, table); char character; // Declare and initialize counters int uppercasecounter = 0; int lowercasecounter = 0; int digitcounter = 0; int wordcounter = 0; int sentencecounter = 0; int ignorecounter = 0; text.get(character); // Input one character do // Process each character ProcessCharacter(character, uppercasecounter, lowercasecounter, digitcounter, sentencecounter, wordcounter, ignorecounter); text.get(character); while (text); PrintTable(table, uppercasecounter, lowercasecounter, digitcounter, sentencecounter, wordcounter, ignorecounter); text.close(); table.close(); return 0; // Function Decode examines the character and returns its type

64 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Features Decode( /* in */ char character ) // Character decoded static bool hyphen = false; if (hyphen) hyphen = false; if (character == '\n') return IGNORE; else return COMPOUND; if (isupper(character)) return UPPER; else if (islower(character)) return LOWER; else if (isdigit(character)) return DIGIT; else switch (character) case '.' : case '?' : case '!' : return EOS; case ' ' : case ',' : case ';' : case ':' : case '\n' : return EOW; case '-' : hyphen = true; return IGNORE; return IGNORE; // Function OpenFiles reads in the names of the input file and the // output file and opens them for processing. void OpenFiles( /* inout */ ifstream& text, // Input file /* inout */ ofstream& table ) // Output file

65 DalePhatANS_complete 8/18/04 10:31 AM Page 1113 Answers to Selected Exercises 1113 string infilename; string outfilename; cout << "Enter the name of the file to be processed" << endl; cin >> infilename; text.open(infilename.c_str()); cout << "Enter the name of the output file" << endl; cin >> outfilename; table.open(outfilename.c_str()); table << "Analysis of characters on input file " << infilename << endl << endl; // Function PrintTable prints the percentages represented by each // of the five categories void PrintTable ( /* inout */ ofstream& table, // Output file /* in */ int uppercasecounter, // Uppercase letters /* in */ int lowercasecounter, // Lowercase letters /* in */ int digitcounter, // Digits /* in */ int sentencecounter, // '.', '?', '!' /* in */ int wordcounter, // Words /* in */ int ignorecounter ) // Everything else int totalalphanum; totalalphanum = uppercasecounter + lowercasecounter + digitcounter; // Print results on file table table << "Total number of alphanumeric characters: " << totalalphanum << endl; table << "Number of uppercase letters: " << uppercasecounter << endl; table << "Number of lowercase letters: " << lowercasecounter << endl; table << "Number of digits: " << digitcounter << endl; table << "Number of characters ignored: " << ignorecounter << endl; wordcounter = wordcounter + sentencecounter; table << "Number of Words: " << wordcounter << endl;

66 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises table << "Number of Sentences: " << sentencecounter << endl; table << "Average word length: " << fixed << setprecision(2) << float(totalalphanum)/wordcounter << endl; table << "Average sentence length: " << fixed \ << setprecision(2) << float(wordcounter) / sentencecounter << endl; // Function ProcessCharacter examines character and increments the // appropriate counter. void ProcessCharacter ( /* in */ char character, // Character to be // processed /* inout */ int uppercasecounter, // Uppercase letters /* inout */ int lowercasecounter, // Lowercase letters /* inout */ int digitcounter, // Digits /* inout */ int sentencecounter, // '.', '?', '!' /* inout */ int wordcounter, // Words /* inout */ int ignorecounter ) // Everything else static bool endofword = false; switch (Decode(character)) case UPPER : uppercasecounter++; endofword = false; break; case LOWER : lowercasecounter++; endofword = false; break; case DIGIT : digitcounter++; endofword = false; break; case EOW : if (!endofword) wordcounter++; endofword = true; break;

67 DalePhatANS_complete 8/18/04 10:31 AM Page 1115 Answers to Selected Exercises 1115 case EOS : sentencecounter++; endofword = true; break; case IGNORE: ignorecounter++; break; case COMPOUND : wordcounter++; switch (Decode(character)) case UPPER : uppercasecounter++; break; case LOWER : lowercasecounter++; break; case DIGIT : digitcounter++; break; 4. You could test to see if it had been set. if (endofword) endofword = false; However, it takes about the same amount of time to just set it to the same value. 5. This program looks at each character in a file and classifies it as an uppercase letter, a lowercase letter, a digit, an end-of-sentence symbol, an end-of-word symbol, or a hyphen in the expanded version in Exercise 3. No errors can occur in classifying each character, so no error detection is necessary. There may be logic errors in the determination of the number of words and sentences, but no error detection would find these. Chapter 11 Exam Preparation Exercises 1. False. 2. True. 3. True. 4. True. 5. yourname myname first middle last first middle last a. George b. George Smith c. George Smith George Smith d. George Smith George Nathaniel Smith e. George N. Smith George Nathaniel Smith

68 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 6. Assignment, passing as a parameter, return value from a function. 7. An enumeration type holds values corresponding to different identifiers. A union holds values of different types at different times. 8. a. sally.studentname.first = "Sally"; sally.studentname.middle = "Ellen"; sally,studentname.last = "Strong"; b. sally.gradenumber = 7; c. spring = sally.grades[3] 9. All of the fields of the struct are copied to the value parameter, and the function works with the copy. Only the address of the struct would be passed to the reference parameter, and the function would have direct access to the fields of the struct. 10. a. Makes the type of grade be char, and inputs a char to grade. b. Compares the value in grade to see if it is in the range of A' through D'. c. Computes the integer equivalent of the letter grade, changes the type of grade to int, and assigns the numeric value to it. 11. Its domain and operations. 12. The data representation is a concrete form of the domain, expressed in terms of structures and types supported by the programming language. 13. Private. 14. current.plus(period); 15. Observers, because they do not change the private data. 16. By matching the number, order, and types of the arguments with the parameter list. 17. Calendar primary; Calendar primary(2004); 18. It is written with a tilde before the name. Chapter struct Time int minutes; int seconds; Programming Warm-Up Exercises 2. sometime.minutes = 6; sometime.seconds = 54; 3. struct Song string title; string album; string artist; Time playtime; Category type;

69 DalePhatANS_complete 8/18/04 10:31 AM Page 1117 Answers to Selected Exercises Song mysong; mysong.title = "Long Run for a Collie Dog"; mysong.album = "Timmy's Nightmare" mysong.artist = "Rebekah MacIntash"; mysong.playtime = sometime; mysong.type = BLUES; 5. cout << mysong.playtime.minutes << ":" << mysong.playtime.seconds; 6. union Temporal string asstring; int asinteger; Time astime; 7. Temporal shift; shift = sometime; 8. //***************************************** // SPECIFICATION FILE (Period) // This file gives the specification of a // class for geologic time periods //***************************************** class Period public: enum PeriodName PRECAMBRIAN, CAMBRIAN, ORDOVICIAN, SILURIAN, DEVONIAN, CARBONIFEROUS, PERMIAN, TRIASSIC, JURASSIC, CRETACEOUS, TERTIARY, QUATERNARY; Period(); // Default constructor // Creates object with value PRECAMBRIAN Period(PeriodName); // Constructor // Creates object with value specified String ToString() const; // Observer // Returns name of object as a string

70 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises int ToInt() const; // Observer // Returns value of object as an int, 0 = PRECAMBRIAN float StartDate() const; // Observer // Returns starting date of period in millions of years ago void Increment(); // Transformer // Advances value of object to next most recent period // Won't advance past QUATERNARY private: PeriodName thisperiod; 9. Period quaternary(quaternary); for (Period count; count.toint() <= quaternary.toint(); count.increment()) cout << count.tostring() << ": " << count.startdate() << endl; 10. Period::Period() // Default constructor // Creates object with value PRECAMBRIAN PeriodName = PRECAMBRIAN; Period::Period(/* in */ PeriodName aperiod) // Constructor // Creates object with value specified thisperiod = aperiod; 11. Period::ToInt() const return int(thisperiod); 12. The file would be called Period.h. 13. #include "Period.h"

71 DalePhatANS_complete 8/18/04 10:31 AM Page 1119 Answers to Selected Exercises //***************************************** // SPECIFICATION FILE (Money) // This file gives the specification of a // class for representing money as dollars // and cents using integers. //***************************************** class Money public: Money(); // Default constructor // Creates object with value $0.00 Money(int, int); // Constructor // Creates object with value specified // in dollars and cents int Dollars() const; // Observer // Returns dollars portion of object as an int int Cents() const; // Observer // Returns cents portion of object as an int float Amount() const; // Observer // Returns money value as a float void Add(Money); // Transformer // Adds value of money object in parameter void Sub(Money); // Transformer // Subtracts value of money object in parameter private: int dollars; int cents;

72 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 15. Money::Money() // Default constructor // Creates object with value $0.00 dollars = 0; cents = 0; Money::Money(/* in */ int userdollars, /* in */ int usercents) // Constructor // Creates object with value specified dollars = userdollars; cents = usercents; 16. The file name would be Money.h. 17. #include "Money.h" 18. float Money::Amount() const // Observer // Returns money value as a float return float(dollars) + float(cents)/100.0; Chapter 11 Case Study Follow-Up Exercises 1. There are two constructors, one default and one parameterized. The other five functions are observer functions. FirstName, MiddleName, and LastName return the value in the class fields. MiddleInitial extracts the first character in the middle name and returns it. ComparedTo takes two instances of the class and returns their relative positions in terms of alphabetic ordering. 2. To test the NameType ComparedTo function, pairs of names must be compared that allow each of the return statements to be executed. Pairs of names satisfying the following conditions should be included: a. The last name is less than the second last name. b. The last name is greater than the second last name. c. The last names are equal, and the first first name is less than the second first name. d. The last names are equal, and the first first name is greater than the second first name. e. The last names and first names are equal, and the first middle name is less than the second middle name. f. The last names and first names are equal, and the first middle name is greater than the second middle name. g. The last names, first names, and middle names are the same.

73 DalePhatANS_complete 8/18/04 10:31 AM Page 1121 Answers to Selected Exercises 1121 It is easier to choose one name and then vary the one to which it is compared. The letter in the Reason for Test Case column refers back to the letters above. The name David John Jones is compared to the input, which is the second name. This is not repeated in the table. The expected output column does not show the completely documented output, just the path. Reason for Input Expected Output Observed Test Case Output a. David John Smith Jones comes before Smith b. David John Allen Jones comes after Allen c. William John Jones David Jones comes before William Jones d. Allen John Jones David Jones comes after Allen Jones e. David William Jones David John Jones comes before David William Jones f. David Allen Jones David John Jones comes after David Allen Jones g. David John Jones They are the same 3. #include <iostream> #include "NameType.h" using namespace std; int main() NameType name1("david", "John", "Jones"); string name1w = name1.firstname() + ' ' + name1.middlename() + ' ' + name1.lastname(); for (int count = 1; count <= 7; count++) NameType name2; string name2w = name2.firstname() + ' ' + name2.middlename() + ' ' + name2.lastname(); switch (name1.comparedto(name2))

74 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises case BEFORE : cout << name1w << " comes before " << name2w << endl; break; case SAME : cout << name1w << " is the same as " << name2w << endl; break; case AFTER : cout << name1w << " comes after " << name2w << endl; return 0;

75 DalePhatANS_complete 8/18/04 10:31 AM Page 1123 Answers to Selected Exercises The initial comes before the same letter in a longer string. The period comes before any letter in a longer string in that position. 5. Enhance the NameType class with a title field and an observer function to return its value. //***************************************************************** // SPECIFICATION FILE (Name.h) // This file gives the specification of the Name abstract // data type. There are two constructors: one takes the first, // middle, and last name as parameters and the second prompts for // and reads the name from the standard input device. //***************************************************************** #include <iostream> #include <string> #include "reltype.h" using namespace std; class Name public: Name(); // Default constructor // first, middle, last, and title have been set to blanks Name( /* in */ string firstname, /* in */ string middlename, /* in */ string lastname, /* in */ string nametitle ); // Parameterized constructor // first is firstname AND middle is middlename AND // last is lastname AND title is nametitle void SetName( /* in */ string firstname, /* in */ string middlename, /* in */ string lastname, /* in */ string nametitle ); // first is firstname AND middle is middlename AND // last is lastname AND title is nametitle

76 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises void ReadName(); // Name is prompted for and read from the standard input // device string FirstName() const; // Return value is this person's first name string LastName() const; // Return value is this person's last name string MiddleName() const; // Return value is this person's middle name char MiddleInitial() const; // Return value is this person's middle initial string Title() const; // PostCondition: // Return value is this person's title RelationType ComparedTo( /* in */ Name othername ) const; // Return value is // BEFORE, if this name comes before othername // SAME, if this name and othername are the same // AFTER, if this name is after othername private: string first; string last; string middle; string title; ; // Person's first name // Person's last name // Person's middle name // Person's title //***************************************************************** // IMPLEMENTATION FILE (Name.cpp) // The file implements the Name member functions. //*****************************************************************

77 DalePhatANS_complete 8/18/04 10:31 AM Page 1125 Answers to Selected Exercises 1125 #include "Name.h" #include <iostream> Name::Name() // Default constructor // first, middle, and last have been set to blanks first = " "; middle = " "; last = " "; title = " "; Name::Name( /* in */ string firstname, // First name /* in */ string middlename, // Middle name /* in */ string lastname, // Last name /* in */ string nametitle) // Title // Parameterized constructor // firstname has been stored in first; middlename has been // stored in middle; lastname has been stored in last; // nametitle has been stored in title first = firstname; // Assign parameters last = lastname; middle = middlename; title = nametitle) void Name::SetName( /* in */ string firstname, // First name /* in */ string middlename, // Middle name /* in */ string lastname, // Last name /* in */ string nametitle) // Title // firstname has been stored in first; middlename has been // stored in middle; lastname has been stored in last; // nametitle has been stored in title

78 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises first = firstname; last = lastname; middle = middlename; title = nametitle; // Assign parameters void Name::ReadName() // Name prompted for and read from the standard input device cout << "Enter first name: "; // Prompt for first name cin >> first; // Get first name cout << "Enter middle name: "; // Prompt for middle name cin >> middle; // Get middle name cout << "Enter last name: "; // Prompt for last name cin >> last; // Get last name cout << "Enter title: "; // Prompt for title cin >> title; // Get title string Name::FirstName() const // Return value is first return first; string Name::LastName() const // Return value is last return last; string Name::MiddleName() const // Return value is middle

79 DalePhatANS_complete 8/18/04 10:31 AM Page 1127 Answers to Selected Exercises 1127 return middle; string Name::Title() const // Return value is title return title; char Name::MiddleInitial() const // Return value is middle initial return middle[0]; RelationType Name::ComparedTo( /* in */ Name othername ) const // Precondition: // Input parameter contains a valid name // Return value is // BEFORE, if this name comes before othername // SAME, if this name and othername are the same // AFTER, if this name is after othername if (last < othername.last) return BEFORE; else if (othername.last < last) return AFTER; else if (first < othername.first) return BEFORE; else if (othername.first < first) return AFTER; else if (middle < othername.middle) return BEFORE;

80 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises else if (othername.middle < middle) return AFTER; else return SAME; If you add the title to the ComparedTo function, the titles would be compared as strings. Mr. would come before Mrs. and Dr. would come before Mr. There would be no semantic sense in such a comparison. Chapter 12 Exam Preparation Exercises 1. False. 2. True. 3. False. 4. True. 5. True. 6. C++ does not report an out-of-bounds access as an error. The program accesses the location preceding the first element of the array, which may be a completely unrelated value in memory. 7. Arrays can be passed as arguments to reference parameters. No other aggregate operations are allowed. 8. The base address is the memory location of the first element of an array. It is the value that is passed to a reference parameter in a function call. 9. A string. 10. The inner loop would increment the column index, keeping the row constant. The outer loop would increment the row index to move to the next row after each row is processed. 11. a. 27, b. 15, c. 100,000, The For loop runs from 1 to 100 instead of 0 to 99. The last iteration will assign zero to an out-of-bounds location. 13. Comparison of arrays is not allowed in C Functions cannot return array types. 15. The rows and columns are reversed in the initializer list. This list fits an array that is [4][3]. However, C++ would still place the 12 values within the array just not in the locations that we would expect by looking at their arrangement in the code. 16. Because the index value is input and then used without any checking, an out-of-bounds array access is likely. 17. In addition to returning the maximum value, the function returns with the array ordered least to greatest. 18. * ** * * * ** * * * ** 19. ***

81 DalePhatANS_complete 8/18/04 10:31 AM Page 1129 Answers to Selected Exercises * * * * * * * * * * * * * 21. a. for (int col = 0; col < 15; col++) cout << examprep[0][col] << " "; b. for (int row = 0; row < 15; row++) cout << examprep[row][0] << " "; c. for (int row = 0; row < 7; row++) for (int col = 0; col < 15; col++) cout << examprep[row][col] << " "; cout << endl; d. for (int row = 11; row >= 0; row--) for (int col = 14; col >= 0; col--) cout << examprep[row][col] << " "; cout << endl; Chapter 12 Programming Warm-Up Exercises 1. a. string toptenlist[10]; b. enum Spectrum RED, ORANGE, YELLOW, BLUE, GREEN, INDIGO, VIOLET; float colormix[7]; c. struct dayrecord int Day; string activity; ; enum DayNames SUNDAY, MONDAY, TUESDAY, WEDNESDAY, THURSDAY, FRIDAY, SATURDAY; dayrecord Calendar[6][7]; 2. typedef float DataSet[5]; DataSet input; DataSet output; DataSet working;

82 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 3. DataSet set[3]; 4. for (int row = 0; row < 3; row++) for (int col = 0; col < 5; col++) set[row][col] = 0.0; 5. bool Equals (/* in */ const DataSet first, /* in */ const DataSet second) 6. bool Equals (/* in */ const DataSet first, /* in */ const DataSet second) for(int index = 0; index < 5; index++) if (first[index]!= second[index]) return false; // Only reaches end of loop if all are equal return true; 7. struct Size int height; int width; ; enum Medium OIL, WATERCOLOR, PASTEL, ACRYLIC, PRINT, COLORPHOTO, BWPHOTO; enum Room MAIN, GREEN, BLUE, NORTH, SOUTH, ENTRY, BALCONY; struct ArtWork string artist; string title; Medium medium; Size size; Room room; float price; typedef ArtWork GalleryList[120]; ArtWork currentlist; int numpieces; 8. a. currentlist[36] b. currentlist[11].title c. currentlist[84].size.width d. currentlist[119].room e. currentlist[77].artist[0]

83 DalePhatANS_complete 8/18/04 10:31 AM Page 1131 Answers to Selected Exercises for (int index = 0; index < numpieces; index++) cout << currentlist[index].artist << " " << currentlist[index].title << " " << currentlist[index].price << endl; 10. float total = 0.0; for (int index = 0; index < numpieces; index++) total = total + currentlist[index].price; 11. for (int index = 0; index < numpieces; index++) if (currentlist[index].room == BLUE) cout << currentlist[index].title << endl; 12. float total = 0.0; for (int index = 0; index < numpieces; index++) if(currentlist[index].medium == OIL) if (currentlist[index].size.width * currentlist[index].size.height > 400) total = total + currentlist[index].price; 13. enum Notes C, CSHARP, D, DSHARP, E, F, FSHARP, G, GSHARP, A, ASHARP, B; float scale[8][12]; 14. for (int octave = 0; octave < 8; octave++) for (Notes note = A; note <= GSHARP; note = Notes(note + 1)) cin >> scale[octave][note]; 15. for (Notes note = A; note <= GSHARP; note = Notes(note + 1)) cout << scale[3][note] << endl; 16. for (int octave = 0; octave < 8; octave++) cout << scale[octave][c] << endl; 17. float humidity[10][52][50][3]; 18. void Reset (float humidity[10][52][50][3]) for (int year = 0; year < 10; year++) for (int week = 0; week < 52; week++) for (int state = 0; state < 50; state++) for (Type type = MAX; type <= AVERAGE; type = Type(type + 1)); humidity = 0.0; 19. struct TimePlace int year; int week; int state;

84 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises float difference; TimePlace MaxSpread(float humidity[10][52][50][3]) TimePlace max; float difference; max.year = 0; max.week = 0; max.state = 0; max.difference = 0.0; for (int year = 0; year < 10; year++) for (int week = 0; week < 52; week++) for (int state = 0; state < 50; state++) difference = humidity[year][week][state][max] humidity[year][week][state][min]; if (difference > max.difference) max.difference = difference; max.year = year; max.week = week; max.state = state; return max; 20. for (int year = 5; year < 10; year++) for (int week = 0; week < 52; week++) cout << humidity[year][week][22][average] << endl; Chapter 12 Case Study Follow-Up Exercises 1. The input file cannot be found, a grade is outside the range of 0 to 100, a grade is negative. 2. It would be more efficient to calculate and print the values without storing them. However, the Print Results module would have two major tasks: calculating values and printing them. It is better style to separate these tasks. 3. When an index has semantic content, the index s role in the problem is more than just as an accessor into an array. The index has meaning within the problem itself. In this problem, a grade was used as an index into the grade s frequency counter. If grade is 60, grades[grade] tells how many times the grade 60 has occurred in the file.

85 DalePhatANS_complete 8/18/04 10:31 AM Page 1133 Answers to Selected Exercises void OpenFiles(ifstream& text, ofstream& outfile) // Function OpenFiles reads in the names of the input file and the // output file, opens them for processing, prompts for a heading // for the output file, reads the heading, and prints // it in the file string infilename; string outfilename; string heading; cout << "Enter the name of the file to be processed" << endl; cin >> infilename; text.open(infilename.c_str()); cout << "Enter the name of the output file" << endl; cin >> outfilename; outfile.open(outfilename.c_str()); cout << "Enter a heading for the output file." << endl; cin >> heading; outfile << heading << endl << endl; 5. It might be slightly better to have the input of the heading in the function that does the printing. Chapter 13 Exam Preparation Exercises 1. Because each component except the first has a unique predecessor, and each component except the last has a unique successor. 2. That all of the components have the same type. 3. a. 20, b. 42, c. 42, d while (index < length && fabs(item data[index]) >= EPSILON) We need to change it because floating point numbers cannot be compared reliably for exact equality. 5. temp = data[value1]; data[value1] = data[value2]; data[value2] = temp; 6. It has identified the smallest value in the portion of the list that remains to be sorted. 7. a. 20, b. 21, c. 42, d True. 9. True Because the values in a sorted list are always kept in order. 12. Two iterations. Middle is initially 7, then location 11 (which holds the 12th value) becomes middle. 13. 'H', 'o', 'p', and '\0'

86 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Chapter 13 Programming Warm-Up Exercises 1. bool Deleted (/* in */ ItemType someitem, /* in */ List oldlist, /* in */ List newlist) return oldlist.ispresent(someitem) &&!newlist.ispresent(someitem); 2. void List::Insert( /* in */ ItemType item) // Inserts item into list without duplication if (!IsPresent(item)) data[length] = item; length++; 3. The List type is only defined to have enough room for 50 items. We must change MAX_LENGTH in the specification file to be at least void DeleteAll( /* in */ ItemType item); // Precondition: // NOT IsEmpty() // && item is assigned // IF item is present in list // All occurrences of item are deleted from list // && Length() == // Length()[at]entry number of occurrences of item // ELSE // List is unchanged 5. void List::DeleteAll( /* in */ ItemType item) // Deletes all occurrences of item from the list int index = 0; while (index < length) if (item == data[index])

87 DalePhatANS_complete 8/18/04 10:31 AM Page 1135 Answers to Selected Exercises 1135 data[index] = data[length 1]; length--; index++; 6. void Replace( /* in */ ItemType item, /* in */ ItemType newitem); // Precondition: // NOT IsEmpty() // && olditem is assigned // && newitem is assigned // IF olditem is present in list // First occurrences of olditem is replaced by newitem // ELSE // List is unchanged 7. void List::Replace( /* in */ ItemType olditem, /* in */ ItemType newitem) // Replaces first occurrence of olditem with newitem int index = 0; while (index < length && olditem!= data[index]) index++; if (index < length) data[index] = newitem; 8. The Insert and BinSearch functions must be changed to enable descending order. We don t have to change IsPresent or Delete, because they both use BinSearch to locate the item. 9. The only changes needed to make the binary search work with descending order are to cause the search area to change to the last half of the current search area when the item is less than the middle value, and vice versa. These changes are noted with comments in the following code: void SortedList::BinSearch( /* in */ ItemType item, /* out */ bool& found, /* out */ int& position ) const int first = 0;

88 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises int last = length; int middle; found = false; while (last >= first &&!found) middle = (first + last) / 2; if (item > data[middle]) // Changed < to > last = middle - 1; else if (item < data[middle]) // Changed > to < first = middle + 1; else found = true; if (found) position = middle; 10. void SortedList::DeleteAll( /* in */ ItemType item) // Deletes all occurrences of item from the list int first; int index = 0; while (index < length && item!= data[index]) index++; first = index; while (index < length && item == data[index]) index++; for (int position = index; position < length; position++) data[first + position - index] = data[position]; length = length (index first); 11. The simplest way to do this is to use the member functions Delete and Insert. We could also combine the code from both of these functions into a single function; however, the savings from that approach would be minimal: We would avoid the decrementing and incrementing of length, and save the overhead of calling the functions. void SortedList::Replace( /* in */ ItemType olditem, /* in */ ItemType newitem) // Replaces an occurrence of olditem with newitem

89 DalePhatANS_complete 8/18/04 10:31 AM Page 1137 Answers to Selected Exercises 1137 int oldlength; oldlength = length; Delete(oldItem); if (oldlength > length) // Only insert if olditem deleted Insert(newItem); 12. SortedList indata; ifstream unsorted; ItemType oneitem; unsorted >> oneitem; while (unsorted &&!indata.isfull()) indata.insert(oneitem); unsorted >> oneitem; 13. void SortedList::FilePrint( /* inout */ ofstream& outfile) const int index; for (index = 0; index < length; index++) outfile << data[index] << endl; Chapter 13 Case Study Follow-Up Exercises 1. There is a personal preference. 2. // Statistics Program // This program calculates the average, high score, low score, // number above the average, and number below the average for // a file of test scores. The List ADT is used // Assumption: File "testscores" is not empty and does not contain // more than MAX_GRADES values. // To save space, we omit from each function the precondition // comments that document the assumptions made about valid input // parameter data. These would be included in a program intended // for actual use. // CalculateAverage, CalculateHighest, and CalculateLowest are // combined into one function. #include <fstream>

90 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises #include <iostream> #include <iomanip> #include "list.h" using namespace std; // Function Prototypes void OpenFiles(ifstream& indata, ofstream& outdata); void InputGrades(List& grades, ifstream& indata); void Calculate(List grades, float& average, int& highest, int& lowest); int CalculateAboveAverage(List grades, float average); int CalculateBelowAverage(List grades, float average); void PrintResults(ofstream& outdata, List grades, float average, int highest, int lowest, int aboveaverage, int belowaverage); int main() List grades; float average; int highest; int lowest; int aboveaverage; int belowaverage; // A list of grades // Average grade // Highest grade // Lowest grade // Number of grades above the average // Number of grades below the average // Declare and open files ifstream indata; ofstream outdata; OpenFiles(inData, outdata); if (!indata // i.outdata ) cout << "Files did not open successfully" << endl; return 1; // Process grades InputGrades(grades, indata); Calculate(grades, average, highest, lowest); aboveaverage = CalculateAboveAverage(grades, average); belowaverage = CalculateBelowAverage(grades, average); PrintResults(outData,grades, average, highest, lowest, aboveaverage, belowaverage);

91 DalePhatANS_complete 8/18/04 10:31 AM Page 1139 Answers to Selected Exercises 1139 indata.close(); outdata.close(); return 0; //***************************************************************** void OpenFiles( /* inout */ ifstream& text, /* inout */ ofstream& outfile ) // Function OpenFiles reads in the names of the input file and // the output file and opens them for processing // Files have been opened AND a label has been written on the // output file string infilename; string outdataname; cout << "Enter the name of the file to be processed" << endl; cin >> infilename; text.open(infilename.c_str()); cout << "Enter the name of the output file" << endl; cin >> outdataname; outfile.open(outdataname.c_str()); // Write label on output outfile << "Grade statistics using the List ADT" << endl << endl; void InputGrades( /* inout */ List& grades, // Grade list /* inout */ ifstream& indata ) // Input file // Grades are input from file indata and inserted into grades. // Precondition: // File is not empty // Each grade in the file has been inserted into list of grades

92 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises int grade; // Read grades and put them into the list indata >> grade; while (indata &&!grades.isfull()) grades.insert(grade); indata >> grade; //***************************************************************** void Calculate( /* in */ List grades, /* out */ float& average, /* out */ int& highest, /* out */ int& lowest ) // This function calculates the average test score // Return value is the average grade int sum = 0; int limit = grades.length(); int grade; highest = 0; lowest = 100; grades.reset(); // limit is the number of grades // Prepare for traversal // Add each grade to sum for (int index = 0; index < limit; index++) grade = grades.getnextitem(); sum = sum + grade; if (grade > highest) highest = grade; if (grade < lowest) lowest = grade; average = float(sum) / float(limit); // Return average

93 DalePhatANS_complete 8/18/04 10:31 AM Page 1141 Answers to Selected Exercises 1141 //***************************************************************** int CalculateAboveAverage ( /* in */ List grades, // List of grades /* inout */ float average ) // Average grade // This function calculates the number of grades above the // average // Return value is the number of grades above average int roundedaverage = (int) (average + 0.5); int limit = grades.length(); // Number of grades int grade; int number = 0; grades.reset(); // Prepare for iteration // Calculate the number of grades above the average for (int index = 0; index < limit; index++) grade = grades.getnextitem(); if (grade > roundedaverage) number++; return number; //***************************************************************** int CalculateBelowAverage ( /* in */ List grades, // List of grades /* inout */ float average ) // Average grade // This function calculates the number of grades below the // average // Return value is the number of grades below average int truncatedaverage = (int) (average); int limit = grades.length(); // Number of grades int grade;

94 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises int number = 0; grades.reset(); // Prepare for iteration // Calculate the number of grades below the average for (int index = 0; index < limit; index++) grade = grades.getnextitem(); if (grade < truncatedaverage) number++; return number; //***************************************************************** void PrintResults( /* inout */ ofstream& outdata, // Output file /* in */ List grades, // Grade list /* in */ float average, // Average /* in */ int highest, // Max grade /* in */ int lowest, // Min grade /* in */ int aboveaverage, // Number above /* in */ int belowaverage ) // Number below // Statistics are printed on file outdata // Precondition: // Output file has been successfully opened // Statistics have been written on outdata, appropriately // labeled outdata << "The number of grades is " << grades.length() << endl; outdata << fixed << setprecision(2) << "The average grade is " << average << endl; outdata << "The highest grade is " << highest << endl; outdata << "The lowest grade is " << lowest << endl; outdata << "The number of grades above the average is " << aboveaverage << endl; outdata << "The number of grades below the average is " << belowaverage << endl;

95 DalePhatANS_complete 8/18/04 10:31 AM Page 1143 Answers to Selected Exercises // Statistics Program // This program calculates the average, high score, low score, // number above the average, and number below the average for // a file of test scores. The List ADT is used // Assumption: File "testscores" is not empty and does not contain // more than MAX_GRADES values. // To save space, we omit from each function the precondition // comments that document the assumptions made about valid input // parameter data. These would be included in a program intended // for actual use. // CalculateAboveAverage and CalculateBelowAverage are combined // into one function. #include <fstream> #include <iostream> #include <iomanip> #include "list.h" using namespace std; // Function Prototypes void OpenFiles(ifstream& indata, ofstream& outdata); void InputGrades(List& grades, ifstream& indata); void Calculate(List grades, float& average, int& highest, int& lowest); void CalculateAboveBelow(List grades, float average, int& aboveaverage, int& belowaverage); void PrintResults(ofstream& outdata, List grades, float average, int highest, int lowest, int aboveaverage, int belowaverage); int main() List grades; float average; int highest; int lowest; int aboveaverage; int belowaverage; // A list of grades // Average grade // Highest grade // Lowest grade // Number of grades above the average // Number of grades below the average // Declare and open files

96 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises ifstream indata; ofstream outdata; OpenFiles(inData, outdata); if (!indata // i.outdata ) cout << "Files did not open successfully" << endl; return 1; // Process grades InputGrades(grades, indata); Calculate(grades, average, highest, lowest); CalculateAboveBelow(grades, average, aboveaverage, belowaverage); PrintResults(outData,grades, average, highest, lowest, aboveaverage, belowaverage); indata.close(); outdata.close(); return 0; //***************************************************************** void OpenFiles( /* inout */ ifstream& text, /* inout */ ofstream& outfile ) // Function OpenFiles reads in the names of the input file and // the output file and opens them for processing // Files have been opened AND a label has been written on the // output file string infilename; string outdataname; cout << "Enter the name of the file to be processed" << endl; cin >> infilename; text.open(infilename.c_str()); cout << "Enter the name of the output file" << endl; cin >> outdataname;

97 DalePhatANS_complete 8/18/04 10:31 AM Page 1145 Answers to Selected Exercises 1145 outfile.open(outdataname.c_str()); // Write label on output outfile << "Grade statistics using the List ADT" << endl << endl; void InputGrades( /* inout */ List& grades, // Grade list /* inout */ ifstream& indata ) // Input file // Grades are input from file indata and inserted into grades. // Precondition: // File is not empty // Each grade in the file has been inserted into list of grades int grade; // Read grades and put them into the list indata >> grade; while (indata &&!grades.isfull()) grades.insert(grade); indata >> grade; //***************************************************************** void Calculate( /* in */ List grades, /* out */ float& average, /* out */ int& highest, /* out */ int& lowest ) // This function calculates the average test score // Return value is the average grade int sum = 0; int limit = grades.length(); int grade; // limit is the number of grades

98 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises highest = 0; lowest = 100; grades.reset(); // Prepare for traversal // Add each grade to sum for (int index = 0; index < limit; index++) grade = grades.getnextitem(); sum = sum + grade; if (grade > highest) highest = grade; if (grade < lowest) lowest = grade; average = float(sum) / float(limit); // Return average //***************************************************************** void CalculateAboveBelow ( /* in */ List grades, // List of grades /* in */ float average, // Average grade /* out */ int& aboveaverage, /* out */ int& belowaverage) // This function calculates the number of grades above the // average // Return value is the number of grades above average int roundedaverage = (int) (average + 0.5); int truncatedaverage = (int) (average); int limit = grades.length(); // Number of grades int grade; aboveaverage = 0 ; belowaverage = 0; grades.reset(); // Prepare for iteration // Calculate the number of grades above the average for (int index = 0; index < limit; index++)

99 DalePhatANS_complete 8/18/04 10:31 AM Page 1147 Answers to Selected Exercises 1147 grade = grades.getnextitem(); if (grade > roundedaverage) aboveaverage++; if (grade < truncatedaverage) belowaverage++; //***************************************************************** int CalculateBelowAverage ( /* in */ List grades, // List of grades /* inout */ float average ) // Average grade // This function calculates the number of grades below the // average // Return value is the number of grades below average int limit = grades.length(); int grade; int number = 0; grades.reset(); // Number of grades // Prepare for iteration // Calculate the number of grades below the average for (int index = 0; index < limit; index++) grade = grades.getnextitem(); return number; //***************************************************************** void PrintResults( /* inout */ ofstream& outdata, // Output file /* in */ List grades, // Grade list /* in */ float average, // Average /* in */ int highest, // Max grade /* in */ int lowest, // Min grade /* in */ int aboveaverage, // Number above /* in */ int belowaverage ) // Number below

100 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises // Statistics are printed on file outdata // Precondition: // Output file has been successfully opened // Statistics have been written on outdata, appropriately // labeled outdata << "The number of grades is " << grades.length() << endl; outdata << fixed << setprecision(2) << "The average grade is " << average << endl; outdata << "The highest grade is " << highest << endl; outdata << "The lowest grade is " << lowest << endl; outdata << "The number of grades above the average is " << aboveaverage << endl; outdata << "The number of grades below the average is " << belowaverage << endl; 4. This is a matter of personal style. Chapter 14 Exam Preparation Exercises 1. a. viii, b. iii, c. ix, d. ii, e. vi, f. v, g. vii, h. i, i. iv. 2. True. 3. False. 4. True. 5. In the base class declaration file? 6. Declaration of a class data member to be an object of the class that we want to compose into the new class being defined. 7. The additional data members are sliced off when the object is copied into the variable. 8. The class declaration for DerivedClass doesn t specify that it is making the members of BaseClass public, so they are not available to the client code. The heading should be changed to: class DerivedClass : public BaseClass 9. a. BaseClass: Private data members are basefield. DerivedClass: Private data members are derivedfield and basefield. b. BaseClass: The member functions can access basefield. DerivedClass: The member functions can only access derivedfield. c. BaseClass: The member functions BaseAlpha and BaseBeta can access each other directly. DerivedClass: The functions DerivedAlpha and DerivedBeta can access each other directly, and they can also access BaseAlpha. d. BaseClass: A client may invoke BaseAlpha. DerivedClass: A client may invoke DerivedAlpha, DerivedBeta, or BaseAlpha.

101 DalePhatANS_complete 8/18/04 10:31 AM Page 1149 Answers to Selected Exercises a. The BaseClass constructor is called first. b. The DerivedClass constructor is called last. 11. Because, in pass-by-value, the data members of the derived class arguments are copied into the corresponding members of the base class object parameter, and if there is no corresponding member in the base class, the argument value is discarded. But in pass-byreference, only the address of the argument object is passed, so nothing needs to be discarded. 12. The virtual designation should be used only with the base class declaration of BaseAlpha. 13. The is-a relationship applies to derived classes and their base classes. The has-a relationship describes composition, where one object has a member that is another object. 14. c. Chapter 14 Programming Warm-Up Exercises 1. class IDScore : public TestScore public: IDScore( /* in */ string name, /* in */ int score), /* in */ int IDNumber); int GetIDNumber() const; private: int studentidnumber; ; 2. #include "testscore.h" #include <iostream> #include <string> using namespace std; TestScore::TestScore( /* in */ string name, /* in */ int score) studentname = name; studentscore = score; TestScore::string GetName() const return studentname;

102 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises TestScore::int GetScore() const return studentscore; 3. #include "idscore.h" #include <iostream> #include <string> using namespace std; IDScore::IDScore( /* in */ string name, /* in */ int score, /* in */ int IDNumber ) : TestScore(name, score) studentidnumber = IDNumber IDScore::int GetIDNumber() const return StudentIDNumber; 4. #include "idscore.h" class Exam public: SetScore( /* in */ int location, /* in */ IDScore score ); IDScore GetScore( /* in */ int location ) const; private: IDScore examlist[100]; 5. #include "exam.h" using namespace std; Exam::SetScore(

103 DalePhatANS_complete 8/18/04 10:31 AM Page 1151 Answers to Selected Exercises 1151 /* in */ int location, /* in */ IDScore score ) examlist[location] = score; Exam::IDScore GetScore( /* in */ int location ) const return examlist[location]; 6. class InternPhone : public Phone public InternPhone ( /* in */ int newcountry, /* in */ int newareacode, /* in */ int newnumber, /* in / PhoneType newtype ); void Write () const; private: int country; 7. using namespace std; InternPhone::InternPhone( ( /* in */ int newcountry, /* in */ int newareacode, /* in */ int newnumber, /* in */ PhoneType newtype) : Phone(newAreaCode, newnumber, newtype) country = newcountry; 8. void InternPhone::Write() const cout << country; Phone::Write(); cout << endl;

104 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 9. The Write function of the Phone class must be declared virtual. class Phone public: virtual void Write() const; ; In the WritePhone function, the parameter must be passed by reference: void WritePhone( /* in */ Phone& somephone ) somephone.write(); 10. class InstallRecord : public Computer public: InstallRecord( /* in */ string newname, /* in */ string newbrand, /* in */ string newmodel, /* in */ int newspeed, /* in */ string newserial, /* in */ int newnumber, /* in */ string newlocation, /* in */ SimpleDate newdate ); void Write() const; string getnewlocation(); SimpleDate getsimpledate(); private: string location; SimpleDate installdate; ; 11. Computer::Computer( /* in */ string newname, /* in */ string newbrand, /* in */ string newmodel, /* in */ int newspeed, /* in */ string newserial, /* in */ int newnumber ) name = newname; brand = newbrand; model = newmodel;

105 DalePhatANS_complete 8/18/04 10:31 AM Page 1153 Answers to Selected Exercises 1153 speed = newspeed; serialnumber = newserial; inventorynumber = newnumber; 12. InstallRecord::InstallRecord( /* in */ string newname, /* in */ string newbrand, /* in */ string newmodel, /* in */ int newspeed, /* in */ string newserial, /* in */ int newnumber, /* in */ string newlocation, /* in */ SimpleDate newdate ) : Computer( newname, newbrand, newmodel, newspeed, newserial, newnumber ) location = newlocation; installdate = newdate; 13. void Computer::Write() const; cout << name << endl; cout << brand << endl; cout << model << endl; cout << speed << endl; cout << serialnumber << endl; cout << inventorynumber << endl; 14. void InstallRecord::Write() const; Computer::Write(); cout << location << endl; installdate.write(); 15. Add the following to the specification file: void ReadTime(); // hours, minutes, seconds and time zone have been prompted // for, read, and set

106 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Add the following to the implementation file: void ExTime::ReadTime() // hours, minutes, and seconds have been prompted for, read, // and stored into hrs, mins, and secs string timezone; Time::ReadTime(); cout << "Please enter time zone: EST, CST, MST, " << "PST, EDT, CDT, MDT, or PDT." << endl; cin >> timezone; if (timezone == "PST") zone = EST; else if (timezone == "MST") zone = CST; else if (timezone == "PST") zone = PST; else if (timezone = "EDT") zone = EDT; else if (timezone == "CDT") zone = CDT; else if (timezone == "MDT") zone = MDT; else zone = PDT; Chapter 14 Case Study Follow-Up Exercises 1. // SPECIFICATION FILE (AptTime.h) // This file gives the specification of a Time ADT with // action responsibilities and knowledge reponsibilities class AptTime public: void Set( /* in */ int hours, /* in */ int minutes ); // Precondition: // 0 <= hours <= 23 && 0 <= minutes <= 59 // hrs == hours && mins == minutes

107 DalePhatANS_complete 8/18/04 10:31 AM Page 1155 Answers to Selected Exercises 1155 void Write() const; // Time has been output in the form HH:MM int Hours() const; // Return value is time int Minutes() const; // Return value is minutes bool Equal( /* in */ AptTime othertime ) const; bool LessThan( /* in */ AptTime othertime ) const; AptTime( /* in */ int inithrs, /* in */ int initmins); // Precondition: // 0 <= inithrs <= 23 AND 0 <= initmins <= 59 // Class object is constructed // AND Time is set according to the incoming parameters AptTime(); // hours and minutes have been set to 0 void ReadTime(); // hours and minutes have been prompted for, read, // and set private: int hrs; int mins; ; // IMPLEMENTATION FILE (AptTime.cpp) // This file implements the AptTime member functions #include "apttime.h" #include <iostream>

108 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises using namespace std; // Private members of class: // int hrs; // int mins; void AptTime::Set( /* in */ int hours, /* in */ int minutes ) // Precondition: // 0 <= hours <= 23 && 0 <= minutes <= 59 // // hrs == hours && mins == minutes hrs = hours; mins = minutes; void AptTime::Write() const // Precondition: // The Set function has been invoked at least once // AptTime has been output in the form HH:MM if (hrs < 10) cout << '0'; cout << hrs << ':'; if (mins < 10) cout << '0'; cout << mins; bool AptTime::Equal( /* in */ AptTime otherapttime ) const // Precondition: // The Set function has been invoked at least once // for both this AptTime and otherapttime

109 DalePhatANS_complete 8/18/04 10:31 AM Page 1157 Answers to Selected Exercises 1157 // Function value == true, if this AptTime equals otherapttime // == false, otherwise return (hrs == otherapttime.hrs && mins == otherapttime.mins); bool AptTime::LessThan( /* in */ AptTime otherapttime ) const // Precondition: // The Set function has been invoked at least once // for both this AptTime and otherapttime // && This AptTime and otherapttime represent AptTimes in the // same day // Function value == true, if this AptTime is earlier // in the day than otherapttime // == false, otherwise return (hrs < otherapttime.hrs hrs == otherapttime.hrs && mins < otherapttime.mins); int AptTime::Hours() const // Return value is hrs return hrs; int AptTime::Minutes() const // Return value is mins return mins;

110 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises void AptTime::ReadTime() // hours, minutes, and seconds have been prompted for, read, // and stored into hrs and mins cout << "Enter hours (<= 23): " << endl; cin >> hrs; cout << "Enter minutes (<= 59): " << endl; cin >> mins; AptTime::AptTime() // Default constructor // Postcondition // hrs and mins have been set to 0 hrs = 0; mins = 0; AptTime::AptTime( /* in */ int hours, /* in */ int minutes ) // Precondition: // 0 <= hours <= 23 && 0 <= minutes <= 59 // hrs == hours && mins == minutes hrs = hours; mins = minutes; 2. Yes, it would be appropriate for active time to inherit from static time, because active time is static time and more. 3. Here are the necessary changes. An enumerated type consisting of AM and PM would have to be defined. The constructors and the Set operation would have to include the AM/PM field. The Equal and LessThan operations would have to include AM/PM in the tests.

111 DalePhatANS_complete 8/18/04 10:31 AM Page 1159 Answers to Selected Exercises // SPECIFICATION FILE (AptTime.h) // This file gives the specification of an AptTime ADT with // action responsibilities and knowledge responsibilities. // A field for AM/PM is included. enum AmPm AM, PM; class AptTimeEx public: void Set( /* in */ int hours, /* in */ int minutes, /* in */ AmPm amorpm ); // Precondition: // 0 <= hours <= 23 && 0 <= minutes <= 59 // hrs == hours && mins == minutes // && timeofday == amorpm void Write() const; // Time has been output in the form HH:MM int Hours() const; // Return value is time int Minutes() const; // Return value is minutes AmPm MorningAfternoon() const; // Return value is AM or PM bool Equal( /* in */ AptTimeEx othertime ) const; bool LessThan( /* in */ AptTimeEx othertime ) const; AptTimeEx( /* in */ int inithrs, /* in */ int initmins, /* in */ AmPm amorpm );

112 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises // Precondition: // 0 <= inithrs <= 23 AND 0 <= initmins <= 59 // Class object is constructed // AND Time is set according to the incoming parameters AptTimeEx(); // hours and minutes have been set to 0 AM void ReadTime(); // hours, minutes, and AM or PM have been prompted // for, read, and set private: int hrs; int mins; AmPm timeofday; ; // IMPLEMENTATION FILE (AptTimeEx.cpp) // This file implements the AptTimeEx member functions #include "apttimeex.h" #include <iostream> using namespace std; // Private members of class: // int hrs; // int mins; void AptTimeEx::Set( /* in */ int hours, /* in */ int minutes, /* in */ AmPm amorpm ) // Precondition: // 0 <= hours <= 23 && 0 <= minutes <= 59 // // hrs == hours && mins == minutes

113 DalePhatANS_complete 8/18/04 10:31 AM Page 1161 Answers to Selected Exercises 1161 hrs = hours; mins = minutes; timeofday = amorpm; void AptTimeEx::Write() const // Precondition: // The Set function has been invoked at least once // AptTimeEx has been output in the form HH:MM:SS if (hrs < 10) cout << '0'; cout << hrs << ':'; if (mins < 10) cout << '0'; cout << mins; if (timeofday == AM) cout << " AM"; else cout << " PM"; bool AptTimeEx::Equal( /* in */ AptTimeEx otherapttimeex ) const // Precondition: // The Set function has been invoked at least once // for both this AptTimeEx and otherapttimeex // Function value == true, if AptTimeEx equalsotherapttimeex // == false, otherwise return (hrs == otherapttimeex.hrs && mins == otherapttimeex.mins && otherapttimeex.timeofday == timeofday);

114 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises bool AptTimeEx::LessThan( /* in */ AptTimeEx otherapttimeex ) const // Precondition: // The Set function has been invoked at least once // for both this AptTimeEx and otherapttimeex // && This AptTimeEx and otherapttimeex represent AptTimeExs in the // same day // Function value == true, if this AptTimeEx is earlier // in the day than otherapttimeex // == false, otherwise if (timeofday == AM && otherapttimeex.timeofday == PM) return true; if (timeofday == PM && otherapttimeex.timeofday == AM) return false; return (hrs < otherapttimeex.hrs hrs == otherapttimeex.hrs && mins < otherapttimeex.mins); int AptTimeEx::Hours() const // Return value is hrs return hrs; int AptTimeEx::Minutes() const // Return value is mins return mins; AmPm AptTimeEx::MorningAfternoon() const // Return value is AM or PM

115 DalePhatANS_complete 8/18/04 10:31 AM Page 1163 Answers to Selected Exercises 1163 return timeofday; void AptTimeEx::ReadTime() // hours, minutes, and seconds have been prompted for, read, // and stored into hrs and mins string daynight; cout << "Enter hours (<= 23): " << endl; cin >> hrs; cout << "Enter minutes (<= 59): " << endl; cin >> mins; cout << "Enter AM or PM" << endl; cin >> daynight; if (toupper(daynight[0]) == 'A') timeofday = AM; else timeofday = PM; AptTimeEx::AptTimeEx() // Default constructor // Postcondition // hrs and mins have been set to 0 and timeofday to AM hrs = 0; mins = 0; timeofday = AM; AptTimeEx::AptTimeEx( /* in */ int hours, /* in */ int minutes, /* in */ AmPm amorpm ) // Precondition: // 0 <= hours <= 23 && 0 <= minutes <= 59 // hrs == hours && mins == minutes && timeofday == amorpm

116 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises hrs = hours; mins = minutes; timeofday = amorpm; 5. #include <iostream> #include "AptTimeEx.h" using namespace std; int main() AptTimeEx time1(8, 30, AM); AptTimeEx time2; time1.write(); cout << endl; time2.write(); cout << endl; for (int count = 1; count < 4; count++) time2.readtime(); if (time1.lessthan(time2)) time1.write(); cout << " is less than "; time2.write(); cout << endl; else time1.write(); cout << " is not less than "; time2.write(); cout << endl; if (time1.equal(time2)) time1.write(); cout << " is equal to "; time2.write(); cout << endl;

117 DalePhatANS_complete 8/18/04 10:31 AM Page 1165 Answers to Selected Exercises 1165 else time1.write(); cout << " is not equal to "; time2.write(); cout << endl; return 0; 6. This is an activity. No answer is expected. Chapter 15 Exam Preparation Exercises 1. a. v, b. iii, c. i, d. iv, e. vi, f. ii. 2. a. ii, b. i, c. iii, d. v, e. vi, f. iv. 3. True. 4. True. 5. False. 6. True. 7. The pointed-to data is returned to the heap. 8. The pointer(s) are deleted but the pointed-to data is not. It thus becomes inaccessible, and contributes to memory leakage. 9. It produces an inaccessible int because the int allocated with the first new operation is not deleted. 10. It creates and references a dangling pointer in germanshorthair. 11. b. 12. b. 13. e (the ++ takes precedence over the *, so the address in finger is incremented). 14. faxlog 15. The shallow copy creates pointers from the new object to the original object s pointed-to data. When that data is deleted, the new object then has one or more dangling pointers. 16. A destructor. 17. False. 18. *libraryrecord Chapter 15 Programming Warm-Up Exercises 1. int* intpointer; int someint; intpointer = &someint; someint = 451; *intpointer = 451;

118 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 2. char* chararrpointer; char[4] initials; chararrpointer = initials; chararrpointer[1] = 'A'; chararrpointer[2] = 'E'; chararrpointer[3] = 'W'; 3. struct Phone int country; int area; int number; Phone newphone; Phone* structpointer; structpointer = &newphone; structpointer->country = 1; structpointer->area = 888; structpointer->number = ; 4. struct Phone int country; int area; int number; Phone newphone; Phone& structreference = newphone; structreference.country = 1; structreference.area = 888; structreference.number = ; 5. bool ShallowCompare ( /* in */ structpointer first, /* in */ structpointer second ) return (first == second); 6. bool DeepCompare ( /* in */ structpointer first, /* in */ structpointer second ) return (first->country == second->country && first->area == second->area && first->number == second->number);

119 DalePhatANS_complete 8/18/04 10:31 AM Page 1167 Answers to Selected Exercises static max = data[0]; for(int index = 1; index < 100; index++) if (data[index] > max) max = data[index]; delete [] data; 8. int Greatest( ( /* inout */ int data[], /* in */ int size ) static max = data[0]; for(int index = 1; index < size; index++) if (data[index] > max) max = data[index]; delete [] data; return max; 9. int GetGreatestInput() int size; int* values; cout << "Enter number of data values: "; cin >> size; values = new int[size]; for(int index = 0; index < size; index++) cout << "Enter integer value: "; cin >> values[index]; return Greatest(values, size); 10. ~Circuit() delete [] source; delete [] sink; 11. if (oldvalue == NULL) newvalue = new int; else newvalue = oldvalue; 12. if (oldvalue!= newvalue) delete oldvalue;

120 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Chapter 15 Case Study Follow-Up Exercises 1. The other functions can be tested within the context of testing the ComparedTo function. Here are the cases that must be addressed. a. The year is less than the second year. b. The year is greater than the second year. c. The years are equal, and the first month is less than the second month. d. The years are equal, and the first month is greater than the second month. e. The years and months are equal, and the first day is less than the second day. f. The years and months are equal, and the first day is greater than the second day. g. The years, months, and days are the same. It is easier to choose one date and then vary the one to which it is compared. The letter in the Reason for Test Case column refers back to the letters above. The date 10/30/2000 is compared to the input, which is the second date. This is not repeated in the table. The output column does not show the completely documented output, just the path. Reason for Input Expected Output Observed Test Case Output a. 9/12/ comes before 2001 b. 10/30/ comes after 1956 c. 11/9/ /2000 comes before 11/2000 d. 9/5/ /2000 comes after 9/2000 e. 10/31/ /30/2000 comes before 10/31/2000 f. 10/4/ /30/2000 comes after 10/4/2000 g. 10/30/2000 They are the same If the test plan seems familiar, look at the solution to Case Study Follow-Up Exercise #include <iostream> #include "Date.h" using namespace std; int main() int month; int day; int year; Date date1; date1.set(10, 30, 2000); cout << "date1 is " << date1.month() << "/" << date1.day() << "/" << date1.year() << endl; Date date2; for (int count = 1; count <= 7; count++)

121 DalePhatANS_complete 8/18/04 10:31 AM Page 1169 Answers to Selected Exercises 1169 cout << "Enter month" << endl; cin >> month; cout << "Enter day" << endl; cin >> day; cout << "Enter year" << endl; cin >> year; date2.set(month, day, year); switch (date1.comparedto(date2)) case BEFORE : cout << "date1 comes before date2" << endl; break; case SAME : cout << "date1 is the same as date2" << endl; break; case AFTER : cout << "date1 comes after date2" << endl; return 0; 2. Copy constructors are needed when initializing a variable in a declaration, when passing a variable as a value parameter, and when returning the variable as the value of a function. None of these cases occurs in the problem for which the SortedList class was implemented; thus, no copy constructor was needed. A destructor is implicitly invoked when the class object is destroyed. In the problem for which we used the SortedList class, the objects were declared in the main function; thus, a destructor would not have been called, and was thus unnecessary. However, it would be safer and better style to include these functions because you can never tell how a user might use objects of this class in their programs. 3. Recall that a top-down design produces a hierarchy of tasks, and an object-oriented design produces a hierarchy of objects. Clearly the process used to build the appointment calendar demonstrated object-oriented design. 4. The next step is to put all the pieces together to form a collection of days ordered by date. The Case Studies in Chapters 12 and 13 demonstrated two different ways to implement a collection: using the data itself as an index and keeping a list of the data. Thus, the collection of days can be implemented as follows: A list of days A one-dimensional array in which the month is used as an index into a list of days for the month A two-dimensional array in which the month is one index and the day is the other index and each cell contains the appointments for one day

122 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Chapter 16 Exam Preparation Exercises 1. a. An individual element of a linked list. b. The members of a node that contain the data stored at that location in the list. c. The member of a node that contains the address of the next node in the list. d. The pointer that keeps track of where processing is taking place in the list. 2. False. 3. True. 4. a. The address of the next node in the list. b. The data in the current node. c. The data in the next node in the list. d. The address of the node after the next node. e. The data in the node after the next node. f. The data in the fourth node in the list. 5. a. It stores the index. b. It stores the address. 6. a. An empty list. b. That processing has reached the end of the list. 7. Search the list to find the node whose successor is greater than the new node. Set the link field of the new node to equal the link field of the currptr node. Set the link field of the currptr node equal to the address of the new node. 8. Set a temporary pointer equal to the link field of the current node, set the current node link field equal to the link field of its successor, and delete the data pointed- to by the temporary pointer. 9. a. Linked, b. Array, c. Array, d. Array. 10. A direct array representation keeps the elements physically adjacent in the array. A linked list implemented with an array allows elements to be in any physical order in the array, and connects them logically using a link field in each element. 11. The typedef refers to NodeType before it has been declared. We need to add a forward declaration of NodeType before the typedef that uses it. 12. a. An array the list is small and bounded, and searches are done frequently. b. A linked list the list varies unpredictably in size, and does not require frequent searching. c. A linked list the list is unbounded, and deletion is at the head. Chapter 16 Programming Warm-Up Exercises 1. a. node[head].component b. currptr = node[currptr].link c. node[node[currptr].link].component d. node[node[node[currptr].link].link].component 2. a. head->component b. currptr = currptr->link c. (currptr->link)->link d. ((currptr->link)->link)->link

123 DalePhatANS_complete 8/18/04 10:31 AM Page 1171 Answers to Selected Exercises head = new NodeType; head->component = 100; currptr = head; 4. newnodeptr = new NodeType; newnodeptr->component = 212; currptr->link = newnodeptr; currptr = newnodeptr; 5. newnodeptr = new NodeType; newnodeptr->component = 32; currptr->link->link = newnodeptr; currptr = newnodeptr; 6. NodePtr saveptr; saveptr = head; head = currptr->link; head->link = saveptr; currptr->link = currptr->link->link; 7. auxptr = currptr; while (auxptr->link!= NULL) auxptr= currptr->link; auxptr->null; 8. auxptr = currptr; if (auxptr!= NULL) if (auxptr->link == NULL) auxptr = NULL; else while (auxptr->link!= NULL) auxptr= currptr->link; auxptr->null; 9. void Sort(NodePtr& head, NodePtr& currptr) NodePtr; // head of sorted list NodePtr lastptr; // pointer to end of new list NodePtr tailptr = NULL; // trailing pointer when searching NodePtr tempptr; // used for traversal NodePtr smallptr; // pointer to smallest NodePtr smalltrailing; // pointer to one before smallest while (head!= NULL) tempptr = head;

124 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises smallptr = head; tailptr = NULL; smalltrailing = NULL; while (tempptr!= NULL) // find smallest if (smallptr->component > tempptr->component) smallptr = tempptr; smalltrailing = tailptr; tailptr = tempptr; tempptr = tempptr->link; // remove node if (smalltrailing == NULL) // only one node left head = smallptr -> link; // set end of loop else smalltrailing->link = smallptr->link; // remove node // insert into new list if (heads == NULL) // first node heads = smallptr; heads->link = NULL; lastptr = heads; else // other nodes lastptr->link = smallptr; lastptr = smallptr; lastptr->link = NULL; head = heads; currptr = heads; 10. CopyReverse(NodeType* head, NodeType*& headr) headr = NULL; // head of new list NodePtr newnode; NodePtr tempptr = head; // used for traversal

125 DalePhatANS_complete 8/18/04 10:31 AM Page 1173 Answers to Selected Exercises 1173 while (tempptr!= NULL) newnode = new NodeType; newnode->component = tempptr->component; newnode->link = headr; headr = newnode; tempptr = tempptr->link Chapter 16 Case Study Follow-Up Exercises 1. No, we could not. The items in the lists implemented thus far are encapsulated; the items cannot be changed. One of the operations required for the list of days is inserting an entry into the appropriate day, thus changing the contents of the day object. 2. void Day::InsertEntry( /* in */ Entry newentry ) // If list is full, EntryListIsfull exception is thrown; if // time is already filled, DuplicateTime exception is thrown; // else newentry is inserted in list if (list.isfull()) throw EntryListIsFull(); else if (list.ispresent(newentry)) throw DuplicateTime(); else list.insert(newentry); 3. #include "Name.h" // ShortName specification class ShortName : public Name public: void ReadName(); ShortName(/* in */ string firstname, /* in */ string middlename, /* in */ string lastname); ShortName(); string FirstName() const; // Return value is this person's first name

126 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises string LastName() const; // Return value is this person's last name private: string first; string last; ; #include "ShortName.h" // ShortName implementation void ShortName::ReadName() // Name prompted for and read from the standard // input device cout << "Enter first and last names " << "separated by a blank." << endl; cin >> first >> last; Name:SetName(first, " ", last); ShortName::ShortName(/* in */ string firstname, /* in */ string middlename, /* in */ string lastname) : Name(firstName, middlename, lastname) ShortName::ShortName() string ShortName::FirstName() const // Return value is this person's first name return Name::FirstName(); string ShortName::LastName() const

127 DalePhatANS_complete 8/18/04 10:31 AM Page 1175 Answers to Selected Exercises 1175 // Return value is this person's last name return Name::LastName(); 4. void AptTime::ReadTime() // hours, minutes, and seconds have been prompted for, read, // and stored into hrs and mins cout << "Enter hours (<= 23)and minutes (<=59) " << "separated by a blank" << endl; cin >> hrs >> mins; 5. This is a matter of taste, but most of you will prefer the shorter entry. 6. There are several ways to look at this question. If we knew for sure that the calendar only contained appointments for one year, then it probably would be more efficient to just check the day; however, there is nothing in the program that assumes that only one year is represented. As it stands now, the calendar can represent any number of years. Thus, using the date comparison takes care of both situations. Chapter 17 Exam Preparation Exercises 1. True. 2. False. 3. Function template. 4. a. Template class, b. Neither (it s a class object) 5. Instantiate To generate a new function (from a function template) or a new class type (from a class template) Template parameter The parameter specified in a function template or class template. Template argument The argument usually a type name that is substituted for a template parameter when a template is instantiated. Specialization A version of a template for a particular template argument. User-defined specialization A template without a template parameter that is an alternative version of an existing template. 6. a. float, b. 3.85, c. Yes. 7. The Try-catch statement. 8. The Throw statement. 9. Each Catch clause must have a formal parameter. 10. a. False. There can be many Catch clauses, as long as their parameters are different. b. True. c. False. A throw statement can appear anywhere. A Throw statement often is located in a function that is called from a Try clause.

128 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 11. a. Control would continue with the statements following the entire Try-catch statement. Almost certainly, this would be wrong. For example, the final comments indicate close the output file, but the output file has already been closed in the Catch clause. b. If we change throw to return, the calling code won t have any idea that a BadData exception occurred. This may or may not be wrong, depending on the design of the program. 12. In the Catch clause, insert return 1; after the output statement. 13. The exception passes to the function that called the function containing the call. If that one also doesn t have a handler for the exception, the process repeats with the next higher caller. If the exception reaches main without finding a handler, it is uncaught; an error message is then output, and the program terminates abnormally. 14. It rethrows the exception. 15. bad_alloc Chapter 17 Programming Warm-Up Exercises 1. template<class SomeType> void PrintSquare(SomeType number) cout << number * number; 2. PrintSquare<int>(10); PrintSquare<long>(10L); PrintSquare<float>(10.0); or PrintSquare(10); PrintSquare(10L); PrintSquare(10.0); 3. template<> void PrintSquare(string value) cout << value << value; 4. a. int Twice( int num ) return 2*num; float Twice( float num ) return 2.0*num; b. cout << Twice(someInt) << ' ' << Twice(someFloat) << endl;

129 DalePhatANS_complete 8/18/04 10:31 AM Page 1177 Answers to Selected Exercises template<class NumType> NumType Sum( NumType arr[], int size ) int i; // Index variable NumType sum = 0; // Sum of the elements for (i = 0; i < size; i++) sum = sum + arr[i]; return sum; 6. template<class SomeType> void GetData( string promptstr, SomeType& data ) cout << promptstr << ' '; cin >> data; 7. template<> void GetData( string promptstr, AutoType& data ) char ch; // The input character cout << promptstr << ' '; cin >> ch; if (ch == 's') data = SEDAN; else if (ch == 'c') data = COUPE; else data = OTHER; 8. a. GList<int> intlist; GList<float> floatlist; b. i = 10; while (i <= 80 &&!intlist.isempty()) // Below, use "while", not "if", because // duplicates are allowed in a list

130 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises while (intlist.ispresent(i)) intlist.delete(i); floatlist.insert(0.5 * float(i)); i++; 9. template<class Type1, class Type2> class MixedPair public: Type1 First() const; Type2 Second() const; void Print() const; MixedPair( Type1 m, Type2 n ); private: Type1 first; Type2 second; ; 10. MixedPair<int, float> pair1(5, 29.48); MixedPair<string, int> pair2("book", 36); 11. template<class Type1, class Type2> MixedPair<Type1, Type2>:: MixedPair( Type1 m, Type2 n ) // Constructor first = m; second = n; template<class Type1, class Type2> Type1 MixedPair<Type1, Type2>::First() const return first; template<class Type1, class Type2> Type2 MixedPair<Type1, Type2>::Second() const return second; template<class Type1, class Type2>

131 DalePhatANS_complete 8/18/04 10:31 AM Page 1179 Answers to Selected Exercises 1179 void MixedPair<Type1, Type2>::Print() const cout << '(' << first << ", " << second << ')' << endl; 12. a. class MathError ; // Don't forget the semicolon b. throw MathError(); // Don't forget the parentheses 13. try str3 = str1 + str2; catch ( length_error ) cout << "*** Error: String too long" << endl; return 1; 14. class SumTooLarge // Exception class ; int Sum( int int1, int int2 ) if (int1 > INT_MAX - int2) throw SumTooLarge(); return int1 + int2; 15. try cout << Sum(oneInt, anotherint) << endl; catch ( SumTooLarge ) cout << "*** Error: Sum is too large" << endl; throw; 16. a. class OpenFailed // Exception class ; void OpenForInput( /* inout */ ifstream& somefile ) // Prompts the user for the name of an input file // and attempts to open the file

132 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises // The user has been prompted for a file name // && IF the file could not be opened // An error message has been printed // && An OpenFailed exception has been thrown string filename; // User-specified file name cout << "Input file name: "; cin >> filename; somefile.open(filename.c_str()); if (!somefile ) cout << "** Can't open " << filename << " **" << endl; throw OpenFailed(); b. Note that the calling code below does not print an error message. The error message has already been printed by OpenForInput, as advertised by its postcondition. ifstream infile; try OpenForInput(inFile); catch ( OpenFailed ) return 1; // Success keep executing Chapter 17 Case Study Follow-Up 1. Duplicate identifiers are not allowed because they are ambiguous. 2. One way to solve the problem would be to put the definition of NodeType in a separate file and have DayList and SortedList2 include that file. Another solution would be to put each definition within a namespace. 3. The test plan used in Chapter 16 should be enhanced with cases to test all of the exceptions. Each of the exception classes must be thrown from each possible place.

133 DalePhatANS_complete 8/18/04 10:31 AM Page 1181 Answers to Selected Exercises 1181 Exceptions Input Values Expected Output Observed Output DuplicateTime DuplicateDay EntryListIsFull UndefinedString UndefinedDay InsertDay InsertApt :30 InsertApt :30 InsertDay InsertApt :30 InsertApt Insertpt InsertApt DeleteApt IsFree PrintDay Entry with same time is already in the list." Day with same date is already defined." Entry list is full for this date." "Input string was not valid." "Read menu and try again." "An operation has been defined on an undefined day" 4. Here are the classes that are changed. //**************************************************************** // SPECIFICATION FILE (AptTime.h) // This file gives the specification of an AptTime ADT with // action responsibilities and knowledge reponsibilities //**************************************************************** #ifndef APT_TIME #define APT_TIME #include "reltype.h" class AptTime public: void Set( /* in */ int hours, /* in */ int minutes ); // Precondition: // 0 <= hours <= 23 && 0 <= minutes <= 59 // hrs == hours && mins == minutes void Write() const; // Time has been output in the form HH:MM int Hours() const; // Return value is time

134 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises int Minutes() const; // Return value is minutes RelationType ComparedTo( /* in */ AptTime othertime ) const; // Precondition: // This and input parameter contains valid times // Return value is // BEFORE, if this time comes before othertime // SAME, if this time and othertime are the same // AFTER, if this time comes after othertime AptTime( /* in */ int inithrs, /* in */ int initmins); // Precondition: // 0 <= inithrs <= 23 AND 0 <= initmins <= 59 // Class object is constructed // AND Time is set according to the incoming parameters AptTime(); // hours and minutes have been set to 0 void ReadTime(); // hours and minutes have been prompted for, read, // and set private: int hrs; int mins; ; #endif // IMPLEMENTATION FILE (AptTime.cpp) // This file implements the AptTime member functions #include "apttimeex.h" #include <iostream>

135 DalePhatANS_complete 8/18/04 10:31 AM Page 1183 Answers to Selected Exercises 1183 using namespace std; // Private members of class: // int hrs; // int mins; void AptTime::Set( /* in */ int hours, /* in */ int minutes ) // Precondition: // 0 <= hours <= 23 && 0 <= minutes <= 59 // // hrs == hours && mins == minutes hrs = hours; mins = minutes; void AptTime::Write() const // AptTime has been output in the form HH:MM if (hrs < 10) cout << '0'; cout << hrs << ':'; if (mins < 10) cout << '0'; cout << mins; // Precondition: // This and input parameter contains valid times // Return value is // BEFORE, if this time comes before othertime // SAME, if this time and othertime are the same // AFTER, if this time comes after othertime

136 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises if (hrs == otherapttime.hrs && mins == otherapttime.mins) return SAME; else if (hrs < otherapttime.hrs hrs == otherapttime.hrs && mins < otherapttime.mins) return BEFORE; int AptTime::Hours() const // Return value is hrs return hrs; int AptTime::Minutes() const // Return value is mins return mins; void AptTime::ReadTime() // hours, minutes, and seconds have been prompted for, read, // and stored into hrs and mins cout << "Enter hours (<= 23): " << endl; cin >> hrs; cout << "Enter minutes (<= 59): " << endl; cin >> mins; AptTime::AptTime() // Default constructor // Postcondition // hrs and mins have been set to 0

137 DalePhatANS_complete 8/18/04 10:31 AM Page 1185 Answers to Selected Exercises 1185 hrs = 0; mins = 0; AptTime::AptTime( /* in */ int hours, /* in */ int minutes ) // Precondition: // 0 <= hours <= 23 && 0 <= minutes <= 59 // && 0 <= seconds <= 59 // hrs == hours && mins == minutes hrs = hours; mins = minutes; //***************************************************************** // SPECIFICATION FILE (Entry.h) // This file contains the specification of the Entry ADT, // which has two contained classes, Name and AptTime //***************************************************************** #ifndef ENTRY #define ENTRY #include "AptTimeEx.h" #include "Name.h" #include <string> #include "reltype.h" using namespace std; class Entry public: string NameStr() const; // Returns a string made up of first name, blank, last name // Return value is first name of Name object, blank, and // last name of Name object string TimeStr() const; // Returns a string made up of hour, colon, minutes

138 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises // Return value is hours from Time object, colon, and // minutes of Time object RelationType ComparedTo(Entry entry) const; // Compares time with entry.time // BEFORE, if this time comes before entry.time // SAME, if this time and entry.time are the same // AFTER, if this time is after entry.time Entry(); // Default constructor // Entry object has been constructed Entry( /* in */ string firstname, // First name /* in */ string middlename, // Middle name /* in */ string lastname, // Last name /* in */ int inithours, // Hours /* in */ int initminutes ); // Minutes // Parameterized constructor // Entry object has been constructed with firstname, // middlename, and lastname as arguments to Name // parameterized constructor; Time object has // inithours and initminutes as arguments // to Time parameterized constructor void ReadEntry(); // Name and time have been prompted for and read private: Name name; AptTime time; ; #endif //***************************************************************** // IMPLEMENTATION FILE (Entry.cpp) // This file contains the specification of the Entry ADT, // which has two contained classes, Name and Time //*****************************************************************

139 DalePhatANS_complete 8/18/04 10:31 AM Page 1187 Answers to Selected Exercises 1187 #include "EntryEx.h" #include <string> #include <fstream> #include <iostream> using namespace std; string Entry::NameStr() const // Returns a string made up of first name, blank, last name // Return value is first name of Name object, blank, and // last name of Name object return (name.firstname() + ' ' + name.lastname()); string Entry::TimeStr() const // Returns a string made up of hour, colon, minutes // Post condition: // Return value is minutes from Time object, colon, and // seconds of Time object ifstream tempin; ofstream tempout; tempout.open("convert"); tempout << time.hours() << ":" << time.minutes() << endl; tempout.close(); tempin.open("convert"); string temp; tempin >> temp; tempin.close(); return temp; void Entry::ReadEntry() // Values have been read from the keyboard into name and time.

140 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises name.readname(); time.readtime(); RelationType Entry::ComparedTo(Entry entry) const // Compares time with entry.time // BEFORE, if this time comes before entry.time // SAME, if this time and entry.time are the same // AFTER, if this time is after entry.time return time.comparedto(entry.time); Entry::Entry() // Default constructor // Entry object has been constructed Entry::Entry( /* in */ string firstname, /* in */ string middlename, /* in */ string lastname, /* in */ int inithours, /* in */ int initminutes) // Constructor initializers : name(firstname, middlename, lastname), time(inithours, initminutes) // Parameterized constructor // Entry object has been constructed with firstname, // middlename, and lastname as arguments to Name // parameterized constructor; inithours, initminutes, // and initseconds as arguments to Time parameterized // constructor // SPECIFICATION FILE (SortedList2.h) // This file gives the specification of a sorted list abstract data // type. The list items are maintained in ascending order of // value

141 DalePhatANS_complete 8/18/04 10:31 AM Page 1189 Answers to Selected Exercises 1189 template<class ItemType> // Forward declaration template<class ItemType> class SortedList2 public: bool IsEmpty() const; // Return value is true if list is empty; // false otherwise bool IsFull() const; // Return value is true if list is full; // false otherwise void Insert( /* in */ ItemType item ); // Precondition: // item is assigned // item is in list // && List components are in ascending order void Delete( /* in */ ItemType item ); // Precondition: // item is somewhere in list // First occurrence of item is no longer in list // && List components are in ascending order void Reset(); // Iteration is initialized ItemType GetNextItem(); // Precondition: // No transformers have been invoked since last call

142 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises // Returns component at the current position // in the SortedList int Length(); // Return value is length of list SortedList2(); // Constructor // Empty list is created SortedList2(int maxlength); // Constructor // Empty list is created SortedList2( const SortedList2& otherlist ); // Copy-constructor // List is created as a duplicate of otherlist ~SortedList2(); // Destructor // List is destroyed bool IsPresent(ItemType item); // Searches the list for item, reporting whether it was found // Return value is true, if item is the list; // false, otherwise private: NodeType<ItemType>* head; NodeType<ItemType>* currentpos; int length; int maxsize; ;

143 DalePhatANS_complete 8/18/04 10:31 AM Page 1191 Answers to Selected Exercises 1191 #include "SortedList2.cpp" // IMPLEMENTATION FILE (SortedList.cpp) // This file implements the SortedList2 class member functions // List representation: a linked list of dynamic nodes #include <iostream> #include <cstddef> // For NULL #include "reltype.h" using namespace std; template<class ItemType> struct NodeType ItemType item; NodeType<ItemType>* link; ; // Private members of class: // NodeType* head; External pointer to linked list template<class ItemType> SortedList2<ItemType>::SortedList2() // Constructor // head == NULL AND maxsize set to 1000 head = NULL; length = 0; maxsize = 1000; template<class ItemType> SortedList2<ItemType>::SortedList2(int maxlength)

144 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises // Constructor // head == NULL head = NULL; length = 0; maxsize = maxlength; template<class ItemType> SortedList2<ItemType>::SortedList2( const SortedList2<ItemType>& otherlist ) // Copy-constructor // IF otherlist.head == NULL (i.e., the other list is empty) // head == NULL // ELSE // head points to a new linked list that is a copy of // the linked list pointed to by otherlist.head TNodeType<ItemType>* fromptr; // Pointer into list being copied TNodeType<ItemType>* toptr; // Pointer into list being built if (otherlist.head == NULL) head = NULL; return; // Copy first node fromptr = otherlist.head; head = new TNodeType<ItemType>; head->item = fromptr->item; // Copy remaining nodes toptr = head; fromptr = fromptr->link;

145 DalePhatANS_complete 8/18/04 10:31 AM Page 1193 Answers to Selected Exercises 1193 while (fromptr!= NULL) toptr->link = new TNodeType<TNodeType>; toptr = toptr->link; toptr->item = fromptr->item; fromptr = fromptr->link; toptr->link = NULL; length = otherlist.length; currentpos = otherlist.currentpos; template<class ItemType> SortedList2<ItemType>::~SortedList2() // Destructor // All linked list nodes have been deallocated from free store TNodeType<ItemType>* temp; // Temporary variable while (!IsEmpty() ) temp = head; head = head->link; delete temp; template<class ItemType> bool SortedList2<ItemType>::IsEmpty() const // Return value is true if head == NULL; false otherwise return (head == NULL);

146 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises //**************************************************************** template<class ItemType> bool SortedList2<ItemType>::IsFull() const // Return value is false return length == maxsize; //**************************************************************** template<class ItemType> void SortedList2<ItemType>::Insert( /* in */ ItemType item ) // Precondition: // item members of list nodes are in ascending order // && item is assigned // New node containing item is in its proper place // in linked list // && item members of list nodes are in ascending order TNodeType<ItemType>* currentptr; // Moving pointer TNodeType<ItemType>* prevptr; // Trailing pointer TNodeType<ItemType>* newnodeptr; // Pointer to new node // Set up node to be inserted newnodeptr = new TNodeType<ItemType>; newnodeptr->item = item; // Find previous insertion point prevptr = NULL; currentptr = head; while (currentptr!= NULL && currentptr->item.comparedto(item) == BEFORE) prevptr = currentptr; currentptr = currentptr->link;

147 DalePhatANS_complete 8/18/04 10:31 AM Page 1195 Answers to Selected Exercises 1195 // Insert new node newnodeptr->link = currentptr; if (prevptr == NULL) head = newnodeptr; else prevptr->link = newnodeptr; length++; template<class ItemType> void SortedList2<ItemType>::Delete( /* in */ ItemType item ) // Precondition: // item members of list nodes are in ascending order // Node containing first occurrence of item is no longer in // linked list // && item members of list nodes are in ascending order TNodeType<ItemType>* currentptr; TNodeType<ItemType>* prevptr; currentptr = head; prevptr = NULL; while (currentptr!= NULL &&!(currentptr->item.comparedto(item) == SAME)) prevptr = currentptr; currentptr = currentptr->link; if (currentptr!= NULL) prevptr->link = currentptr->link; length--;

148 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises template<class ItemType> void SortedList2<ItemType>::Reset() // Iteration is initialized currentpos = NULL; template<class ItemType> ItemType SortedList2<ItemType>::GetNextItem() // Precondition: // No transformers have been invoked since last call // Returns item at the current position in the SortedList and // resets current to next position or first position if // last item is returned ItemType item; if (currentpos == NULL) currentpos = head; item = currentpos->item; currentpos = currentpos->link; return item; template<class ItemType> int SortedList2<ItemType>::Length() // Return value is length return length; template<class ItemType> bool SortedList2<ItemType>::IsPresent(ItemType item) // Searches the list for item, reporting whether it was found // Return value is true if item is the list; // false otherwise

149 DalePhatANS_complete 8/18/04 10:31 AM Page 1197 Answers to Selected Exercises 1197 TNodeType<ItemType>* currentptr; // Search for item // Moving pointer currentptr = head; while (currentptr!= NULL && currentptr->item.comparedto(item)!= SAME) currentptr = currentptr->link; return!(currentptr == NULL); 5. The output should be the same as the output in the text. Chapter 18 Exam Preparation Exercises 1. b. looping. 2. True. 3. True. 4. False. Tail recursion occurs when nothing is done after the return from the recursive call. 5. F(4) = 0 and F(6) = 0. F(5) is undefined. 6. F(3) = 8, F(4) = 16, F(5) = Infinite recursion. 8. It will run out of space on the run time stack. 9. True. 10. c. The value in the variable. 11. c. The distance to the end of the structure abcc d e 15. abcd e f

150 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Chapter 18 Programming Warm-Up Exercises 1. int DigitSum(int number) if (number > 0) return number%10 + DigitSum(number/10); else return 0; 2. int OneDigit (int number) if (number <= 9) return number; else return OneDigit(DigitSum(number)); 3. int Search(const int[] data, value, max, min) int mid; if (max = min) if (data[max] = value) return max; else return -1; else mid = (max + min) / 2; if (data[mid] > value) return Search(data, value, mid-1, min); else if (data[mid] < value) return Search(data, value, max, mid+1); else return mid 4. void Ex4() int number; cout << "Enter a positive number, 0 to end: "; cin >> number;

151 DalePhatANS_complete 8/18/04 10:31 AM Page 1199 Answers to Selected Exercises 1199 if (number!= 0) cout << number << endl; Ex4(); cout << number << endl; 5. void Ex5(int sum) int number; cout << "Enter a positive number, 0 to end: "; cin >> number; if (number!= 0) cout << "Total: " << sum + number <<endl; Ex5(sum + number); cout << number << endl; 6. void Ex6(int sum, int& revsum) int number; cout << "Enter a positive number, 0 to end: "; cin >> number; if (number!= 0) cout << "Total: " << sum + number <<endl; Ex6(sum + number, revsum); revsum = revsum + number; cout << number << " Total: " << revsum << endl; 7. void Ex7(int& greatest) int number; cout << "Enter a positive number, 0 to end: "; cin >> number; if (number > greatest) greatest = number; if (number!= 0)

152 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises Ex7(greatest); cout << number << endl; The greatest must be written immediately following the call. 8. void Ex8(int& greatest) int number; cout << "Enter a positive number, 0 to end: "; cin >> number; if (number > greatest) greatest = number; if (number!= 0) cout << "Greatest: " << greatest << endl; Ex8(greatest); cout << number << endl; The greatest must be written immediately following the call. 9. #include <iostream> using namespace std; void Ex9(int&); int main() int sum = 0; int revsum = 0; Ex9(revSum); cout << "The greatest is " << revsum; return 0; //******************* void Ex9(int& greatest)

153 DalePhatANS_complete 8/18/04 10:31 AM Page 1201 Answers to Selected Exercises 1201 int number; int greatestsofar = 0; cout << "Enter a positive number, 0 to end: "; cin >> number; if (number > 0) Ex9(greatestSoFar); if (number > greatestsofar) greatest = number; else greatest = greatestsofar; cout << number << " Greatest: " << greatest << endl; 10. void CopyReverse(PtrType head1, PtrType& head2) // Assumption: head2 is originally NULL PtrType tempptr; if (head1!= NULL) tempptr = new NodeType; tempptr->info = head1->info; tempptr->link = head2; head2 = tempptr; CopyReverse(head1->link, head2); 11. void Copy(PtrType head1, PtrType& head2) // Assumption: head2 is originally NULL PtrType tempptr; if (head1!= NULL) tempptr = new NodeType; tempptr->info = head1->info; tempptr->link = head2; head2 = tempptr; Copy(head1->link, head2->link);

154 DalePhatANS_complete 8/18/04 10:31 AM Page Answers to Selected Exercises 12. void CopyDouble(PtrType head1, PtrType& head2) // Assumption: head2 is originally NULL PtrType tempptr; if (head1!= NULL) tempptr = new NodeType; tempptr->info = head1->info; tempptr->link = head2; head2 = tempptr; CopyReverse(head1->link, head2); if (head1!= NULL) tempptr = new NodeType; tempptr->info = head1->info; tempptr->link = head2; head2 = tempptr; Copy(head1->link, head2->link); or void CopyDouble(PtrType head1, PtrType& head2) CopyReverse(head1, head2); Copy(head1, head2); Chapter 18 Case Study Follow-Up Exercises 1. The algorithm still works correctly, but because of the lopsided splits it is not so quick. Is this situation likely to occur? That depends on how we choose our splitting value and on the original order of the data in the array. If we use data[first] as the splitting value and the array is already sorted, then every split is lopsided. One side contains one element, while the other side contains all but one of the elements. Thus, our QuickSort is not a quick sort. This splitting algorithm favors an array in random order. 2. The middle value might be a good splitting value: data[(first + last)/2]. 3. Exchange the values data[0] and data[(first + last)/2] at the beginning of each call. Then the same algorithm can be used.