Homework 1 Answers. Due Date: Friday, April 12, 2002 Points: 100

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1 Homework 1 Answers Due Date: Friday, April 12, 2002 Points: 100 UNIX System 1. (10 points) Give a one-line UNIX command that capitalizes the vowels (the letters a, e, i, o, and u ) in a file. Answer: The simplest answer uses the command tr(1): tr aeiou AEIOU 2. (10 points) Use a UNIX program to compute the remainder of 7358^349 when divided by Answer: The program dc(1) is an infinite-precision calculator. Entering the following command: gives the answer, 33. C Programming ^ 1345 % p 3. (25 points) Write a program that asks the user to enter two integers, then calculates and displays their greatest common divisor (GCD). The input and output must look like this: Enter two integers: The GCD of 14 and 49 is 7. where 14 and 49 are numbers the user types and 7 is the corresponding GCD. Be sure to check the input for validity, and terminate the program when the user types an end of file character! (Hint: The classic algorithm for computing the GCD, known as Euclid s Algorithm, goes as follows. Let m and n be variables containing the two numbers. Divide m by n. Save the divisor in m, and save the remainder in n. If n is 0, stop; m contains the GCD. Otherwise, repeat the process, starting with the division of m by n.) Answer: The following program is my solution. Yours may differ! gcd -- compute the GCD of pairs of integers History 1.0 Matt Bishop; original program #include <stdio.h> macros #define BAD_GCD-1 error in arguments -- MUST be non-positive This function returns the greatest common divisor of its arguments Notes: (1) if m = n = 0, the GCD is undefined -- so we return BAD_GCD (2) if m < 0 or n < 0, then gcd(m,n) > 0; so we can just make m and n both positive (3) if m = 0 and n!= 0, gcd(m,n) = n (and vice versa) int gcd(int m, int n) Version of May 6, :03 am Page 1 of 6

2 int rem; remainder for Euclid's algorithm special cases error check -- if both 0, undefined if (m == 0 && n == 0) return(bad_gcd); make all negatives positive if (m < 0) m = -m; if (n < 0) n = -n; now apply Euclid's algorithm while(n > 0) rem = m % n; m = n; n = rem; got it return(m); the main routine int m, n; numbers to take the GCD of int g; the GCD of m and n loop, asking for numbers and printing the GCD while(printf("enter two numbers: "), scanf("%d %d", &m, &n)!= EOF) print the result -- note that if the input is invalid, gcd() simply returns BAD_GCD printf("the GCD of %d and %d is ", m, n); if ((g = gcd(m, n)) == BAD_GCD) printf("undefined.\n"); else printf("%d.\n", g); clean up output and exit putchar('\n'); Version of May 6, :03 am Page 2 of 6

3 4. (30 points) Every computer is limited in the amount of precision it can represent for floating-point numbers. At some point, where epsilon is very small, the following expression will be true: 1.0 == epsilon Write a program to find the largest value of epsilon on your computer. Note that the value of epsilon may be different for floats and doubles. Find both values (and the value for long doubles if your compiler supports them). Also, use 1.0 rather than 0.0 to test epsilon because most computers have special hardware instructions for handling zero arithmetic. (Hint: for gcc(1), a long double is the same as a double.) Answer: This program prints epsilon for both floats and doubles. eps -- a program for determining the largest number d such that 1.0 == d and f such that 1.0F == 1.0F + f History 1.0 Matt Bishop; original version float f; the floating point number we're looking for double d; the double precision number we're looking for loop, dividing the number by 2, until it's negligable then print it for(f = 1.0F; 1.0F!= 1.0F + f; f /= 2) null ; d = f; printf requires a double, not a float printf("for float, epsilon is %g\n", d); loop, dividing the number by 2, until it's negligable then print it for(d = 1.0; 1.0!= d; d /= 2) null ; printf("for double, epsilon is %g\n", d); bye-bye! For both floats and doubles on the CSIF Linus systems, epsilon is Debugging 5. (25 points) On the class web page is the source for a C program named vis.c that reads characters from the standard input and prints them on the standard output after expanding any non-printing characters to their C character escape sequence. Version of May 6, :03 am Page 3 of 6

4 The relevant characters, and the C escape sequences to be printed when those characters are encountered, are: character print as character print as newline \n backslash \\ horizontal tab \t vertical tab \v backspace \b carriage return \r form feed \f bell \a NUL \0 anything else \ooo The anything else entry means that any non-printing character other than the ones named in the table is to be printed as a sequence of three octal digits preceded by a backslash. When the escape sequence for a newline is printed, the program is to skip to the next line. Unfortunately, the program as saved in vis.c will not even compile, let alone run. And the programmer thoughtlessly left off all the comments. Hence, your mission: comment the program, and fix it so it works as described above! You are to turn in a corrected source program, with comments describing the changes you made to get it to work. Answer: The following is the commented, debugged program. Visualize Non-printing Characters Prints all non-printing ASCII characters as C escapes, or (if no such escape), as a 3-digit octal number; also, prints \ as \\. This is useful if you have a file with non-printing chars in it, and need to find out what they are. Note the compiler decides what prints as \a \b \f \n \r \t \v \0 and what prints as a 3-digit octal number Version 1.0 Matt Bishop (bishop@cs.ucdavis.edu) #include <stdio.h> #include <ctype.h> this is the main routine arguments: ignored return: exits with 0 function: print the non-printing and escape characters as visible ASCII sequences exceptions: none int ch; input character do the input a character at a time, Version of May 6, :03 am Page 4 of 6

5 expanding chars using the escape sequences of C; note all are nonprinting EXCEPT for the escape, \ while((ch = getchar())!= EOF) BUG #1 here's the original line if (isprint(ch)) now, as \ is printing, it prints on this branch of the "if" as \ and not in the switch in the else part, where it would print as \\ if (ch!= '\\' && isprint(ch)) putchar(ch); else switch (ch) case '\t': printf("\\t"); break; case '\b': printf("\\b"); break; case '\f': printf("\\f"); break; case '\0': printf("\\0"); break; case '\\': printf("\\\\"); break; case '\v': case '\r': case '\a': case '\n': default: printf("\\v"); break; printf("\\r"); break; BUG #2 you need a : after the '\a' case '\a' printf("\\a"); break; printf("\\a"); break; printf("\\n\n"); break; printf("\\%03o", ch); break; exit gracefully Extra Credit 6. (10 points) In question 4, the hint says that a long double and a double use the same number of bits. Write a program to demonstrate that the size of a long double is the same as that of a double. Answer: The following program uses the sizeof operator to report the number of bytes in each type. Print size of data types Print the number of bytes in a float, a double, and a long double Version 1.0 Matt Bishop (bishop@cs.ucdavis.edu) #include <stdio.h> Version of May 6, :03 am Page 5 of 6

6 this is the main routine arguments: ignored return: exits with 0 function: prints the number of bytes in float, double, long double exceptions: none print the three sizes printf("float is %d bytes\n", sizeof(float)); printf("double is %d bytes\n", sizeof(double)); printf("long double is %d bytes\n", sizeof(long double)); bye! On the CSIF Linux systems, floats are 4 bytes, longs are 8 bytes, and long doubles are 12 bytes. So the hint was wrong. Version of May 6, :03 am Page 6 of 6

7 Homework 1 Answers Due Date: Friday, April 12, 2002 Points: 100 UNIX System 1. (10 points) Give a one-line UNIX command that capitalizes the vowels (the letters a, e, i, o, and u ) in a file. Answer: The simplest answer uses the command tr(1): tr aeiou AEIOU 2. (10 points) Use a UNIX program to compute the remainder of 7358^349 when divided by Answer: The program dc(1) is an infinite-precision calculator. Entering the following command: gives the answer, 33. C Programming ^ 1345 % p 3. (25 points) Write a program that asks the user to enter two integers, then calculates and displays their greatest common divisor (GCD). The input and output must look like this: Enter two integers: The GCD of 14 and 49 is 7. where 14 and 49 are numbers the user types and 7 is the corresponding GCD. Be sure to check the input for validity, and terminate the program when the user types an end of file character! (Hint: The classic algorithm for computing the GCD, known as Euclid s Algorithm, goes as follows. Let m and n be variables containing the two numbers. Divide m by n. Save the divisor in m, and save the remainder in n. If n is 0, stop; m contains the GCD. Otherwise, repeat the process, starting with the division of m by n.) Answer: The following program is my solution. Yours may differ! gcd -- compute the GCD of pairs of integers History 1.0 Matt Bishop; original program #include <stdio.h> macros #define BAD_GCD-1 error in arguments -- MUST be non-positive This function returns the greatest common divisor of its arguments Notes: (1) if m = n = 0, the GCD is undefined -- so we return BAD_GCD (2) if m < 0 or n < 0, then gcd(m,n) > 0; so we can just make m and n both positive (3) if m = 0 and n!= 0, gcd(m,n) = n (and vice versa) int gcd(int m, int n) Version of May 6, :03 am Page 1 of 6

8 int rem; remainder for Euclid's algorithm special cases error check -- if both 0, undefined if (m == 0 && n == 0) return(bad_gcd); make all negatives positive if (m < 0) m = -m; if (n < 0) n = -n; now apply Euclid's algorithm while(n > 0) rem = m % n; m = n; n = rem; got it return(m); the main routine int m, n; numbers to take the GCD of int g; the GCD of m and n loop, asking for numbers and printing the GCD while(printf("enter two numbers: "), scanf("%d %d", &m, &n)!= EOF) print the result -- note that if the input is invalid, gcd() simply returns BAD_GCD printf("the GCD of %d and %d is ", m, n); if ((g = gcd(m, n)) == BAD_GCD) printf("undefined.\n"); else printf("%d.\n", g); clean up output and exit putchar('\n'); Version of May 6, :03 am Page 2 of 6

9 4. (30 points) Every computer is limited in the amount of precision it can represent for floating-point numbers. At some point, where epsilon is very small, the following expression will be true: 1.0 == epsilon Write a program to find the largest value of epsilon on your computer. Note that the value of epsilon may be different for floats and doubles. Find both values (and the value for long doubles if your compiler supports them). Also, use 1.0 rather than 0.0 to test epsilon because most computers have special hardware instructions for handling zero arithmetic. (Hint: for gcc(1), a long double is the same as a double.) Answer: This program prints epsilon for both floats and doubles. eps -- a program for determining the largest number d such that 1.0 == d and f such that 1.0F == 1.0F + f History 1.0 Matt Bishop; original version float f; the floating point number we're looking for double d; the double precision number we're looking for loop, dividing the number by 2, until it's negligable then print it for(f = 1.0F; 1.0F!= 1.0F + f; f /= 2) null ; d = f; printf requires a double, not a float printf("for float, epsilon is %g\n", d); loop, dividing the number by 2, until it's negligable then print it for(d = 1.0; 1.0!= d; d /= 2) null ; printf("for double, epsilon is %g\n", d); bye-bye! For both floats and doubles on the CSIF Linus systems, epsilon is Debugging 5. (25 points) On the class web page is the source for a C program named vis.c that reads characters from the standard input and prints them on the standard output after expanding any non-printing characters to their C character escape sequence. Version of May 6, :03 am Page 3 of 6

10 The relevant characters, and the C escape sequences to be printed when those characters are encountered, are: character print as character print as newline \n backslash \\ horizontal tab \t vertical tab \v backspace \b carriage return \r form feed \f bell \a NUL \0 anything else \ooo The anything else entry means that any non-printing character other than the ones named in the table is to be printed as a sequence of three octal digits preceded by a backslash. When the escape sequence for a newline is printed, the program is to skip to the next line. Unfortunately, the program as saved in vis.c will not even compile, let alone run. And the programmer thoughtlessly left off all the comments. Hence, your mission: comment the program, and fix it so it works as described above! You are to turn in a corrected source program, with comments describing the changes you made to get it to work. Answer: The following is the commented, debugged program. Visualize Non-printing Characters Prints all non-printing ASCII characters as C escapes, or (if no such escape), as a 3-digit octal number; also, prints \ as \\. This is useful if you have a file with non-printing chars in it, and need to find out what they are. Note the compiler decides what prints as \a \b \f \n \r \t \v \0 and what prints as a 3-digit octal number Version 1.0 Matt Bishop (bishop@cs.ucdavis.edu) #include <stdio.h> #include <ctype.h> this is the main routine arguments: ignored return: exits with 0 function: print the non-printing and escape characters as visible ASCII sequences exceptions: none int ch; input character do the input a character at a time, Version of May 6, :03 am Page 4 of 6

11 expanding chars using the escape sequences of C; note all are nonprinting EXCEPT for the escape, \ while((ch = getchar())!= EOF) BUG #1 here's the original line if (isprint(ch)) now, as \ is printing, it prints on this branch of the "if" as \ and not in the switch in the else part, where it would print as \\ if (ch!= '\\' && isprint(ch)) putchar(ch); else switch (ch) case '\t': printf("\\t"); break; case '\b': printf("\\b"); break; case '\f': printf("\\f"); break; case '\0': printf("\\0"); break; case '\\': printf("\\\\"); break; case '\v': case '\r': case '\a': case '\n': default: printf("\\v"); break; printf("\\r"); break; BUG #2 you need a : after the '\a' case '\a' printf("\\a"); break; printf("\\a"); break; printf("\\n\n"); break; printf("\\%03o", ch); break; exit gracefully Extra Credit 6. (10 points) In question 4, the hint says that a long double and a double use the same number of bits. Write a program to demonstrate that the size of a long double is the same as that of a double. Answer: The following program uses the sizeof operator to report the number of bytes in each type. Print size of data types Print the number of bytes in a float, a double, and a long double Version 1.0 Matt Bishop (bishop@cs.ucdavis.edu) #include <stdio.h> Version of May 6, :03 am Page 5 of 6

12 this is the main routine arguments: ignored return: exits with 0 function: prints the number of bytes in float, double, long double exceptions: none print the three sizes printf("float is %d bytes\n", sizeof(float)); printf("double is %d bytes\n", sizeof(double)); printf("long double is %d bytes\n", sizeof(long double)); bye! On the CSIF Linux systems, floats are 4 bytes, longs are 8 bytes, and long doubles are 12 bytes. So the hint was wrong. Version of May 6, :03 am Page 6 of 6