How To Convert Hexadecimal To Binary In Assembly Language

Transcription

1 Arithmetic operations in assembly language Prof. Gustavo Alonso Computer Science Department ETH Zürich Binary! You are probably familiar with decimal and hexadecimal systems. Internally, a CPU works only with binary ( or )! With N bits, one can code different states! Why binary, hexadecimal, octal?! To represent a decimal character, we need 4 bits (,,, 9 = different states) but not all possible combinations are used, i.e., in binary, decimal cannot be efficiently represented.! For hexadecimal characters, we also need 4 bits (,, 2,, E, F = 6 different states) and all combinations are used (the representation is space efficient)! For octal, we need 3 bits (,, 6, 7 = 8 states) and the representation is compact. But we cannot efficiently store octal in bytes and words. 2 N! To calculate a number represented in a system with base B: n 2! where B = base s = sign n = number of digits! The digits always take values in the range [, B-]! The base is often used as subscript to indicate the system, e.g., 7 8 Gustavo Alonso, ETH Zürich. Programming in Assembly 2 d n d... d2dd n N = s i d B i Conversions between systems! Conversion from decimal to binary through division by powers of 2: 374 >= 256 yes = rest =8 8 >= 28 no = 8 >= 64 yes = rest 8-64=54 54 >= 32 yes = rest 54-32=22 22 >= 6 yes = rest 22-6=6 6 >= 8 no = 6 >= 4 yes = rest 6-4=2 2 >= 2 yes = rest 2-2= >= no = RESULT! As alternative, one can take the rest of successive divisions by 2: /2 = 87 87/2 = 93 93/2 = 46 46/2 = 23 23/2 = /2 = 5 5/2 = 2 2/2 = /2 =! The same can be done using octal and then transforming to binary 374/8 = 46 6 = 46/8 = 5 6 = 6/8= 6 5 = octal 566 = binary! Conversion to hexadecimal can be done by converting to binary and then grouping bits 4 by 4 Gustavo Alonso, ETH Zürich. Programming in Assembly 3 Coding characters! ASCII character set (American standard code for information interchange) represents characters with 7 bits. Often also written as 2 or 3 octal digits! In C, it corresponds to the types char or unsigned char! In Assembly can be coded directly with ` oder " or with octal mov 4, %r mov a, %r mov "a", %r! Anything beyond this has coding and standardization problems "Chinese has over 42' symbols/characters! There are also competing standards for 8 bit character sets: "Code Page 85 "ISO " MacCharacterset Gustavo Alonso, ETH Zürich. Programming in Assembly 4

2 Boolean operations! With 2 bits: A B and or nand xor! NOT can be simulated not source_reg, dest_reg xnor source_reg, %g, dest_reg! Synthetic operations such as mov are also implementable with boolean operations or %g, source_reg, dest_reg! or clr (clear): or %g, %g, dest_reg! All possible operations Test operation! Imagine you have to write in assembly something like: if (a>) b++;! A first approach could be (assume a is in %l, and b in %l2): next: cmp %l, ble next nop %l2,, %l2! translating the cmp (which is a synthetic operation) we get: next: subcc %l, %g, %g! sets CC ble next nop %l2,, %l2! An alternative way is to use the tst instruction (test): next: tst %l ble next nop %l2,, %l2! tst is synthetic and is translated as next: orcc %l, %g, %g! sets CC ble next nop %l2,, %l2! All these testing instructions affect the condition code bits Gustavo Alonso, ETH Zürich. Programming in Assembly 5 Gustavo Alonso, ETH Zürich. Programming in Assembly 6 Setting, clearing and testing flags! Bits are often used to represent boolean flags. To manipulate these flags, the assembly language provides a number of synthetic operations bset bclr btog = or (bit set) = andn (bit clear) = xor (bit toggle)! To check whether one or more flags are set, btst is used. For instance, to check whether flags x and x8 are set, we use: btst x8, %l! which is translated into andcc %l, x8, %g! This is one of the principles of RISC architectures: use very simple, very efficient building blocks. More complicated or fancier operations are not implemented by the processor but translated by the assembler into the simple instructions supported by the processor.! Do not forget that the only reason why we need more complicated and fancier instructions is for the convenience of the programmer Adding numbers in binary! Remember how to two numbers in the decimal system? sum 6 carry! the carry value is moved to the left column and ed to the digits there: sum 9 carry!...! When we want to two binary numbers of a single bit each, ition is done in a similar manner: A B sum carry! The interesting aspect is that the sum is the same as an exclusive or (XOR) and the carry is the same as an AND. Hence, we can implement ition of binary numbers as a sequence or XOR and AND operations Gustavo Alonso, ETH Zürich. Programming in Assembly 7 Gustavo Alonso, ETH Zürich. Programming in Assembly 8

3 Addition as a sequence of AND and XOR! The following C program illustrates how this works: (int a, int b){ int s,c; /* s is the sum, c is the carry */ Addition in hardware s = a^b; /* sum is xor */ while (c = (a&b) << ){ a = s; b = c; s = a^b; } return(s); } /* (c = (a&b) << ) means: c = a AND b, shift the bits of c one position to the left */ Gustavo Alonso, ETH Zürich. Programming in Assembly 9 Gustavo Alonso, ETH Zürich. Programming in Assembly Modulus arithmetic! In computers, the size of a numbers is limited by the size of the storage space dedicated to it! Because of the storage constraints, computers often work with modulus arithmetic! Modulus arithmetic considers only numbers in a range [,M), that is, a number n must be such that <= n < M. M is the modulus! In modulus arithmetic, once we reach the largest possible number (M-) we start again from the beginning ()! When we perform an operation on two numbers and the results exceeds the modulus, we say that an overflow has occurred. For instance, if we work modulus 6 in binary arithmetic and we + 7: = 7 = + result = The result can only be interpreted in terms of Modulus arithmetic: + 7 = 8 = Gustavo Alonso, ETH Zürich. Programming in Assembly Complements! Modulus arithmetic gives a significant degree of flexibility when working with numbers.! In modulus, we can use modulus arithmetic to represent negative numbers: ! We will use complements to implement ion! Complement is important because it can be used to turn ion into a simple sum: a-b = a + ( n r - - b) +! where n- - b is the diminished radix complement, r! and - - b + is the radix n complement r for numbers with n digits using a system with base r! For example, in decimal (r=) and with 2 digits (n=2) 23-7 = 23+( - - 7) + = 23+(99-7) 2 + = = = 6! and 93 is the radix complement of 7 Gustavo Alonso, ETH Zürich. Programming in Assembly 2

4 Subtraction by complementing! Let b be a binary number of, e.g., 4 bits:! 2 4 b is the one s complement of number b (same as ing from all s)! 2 4 b + is the two s complement of number b (is the one s complement plus one)! To perform a - b, we a and the two s complement of b: = 2 = two s complement(2) = + = 2! This also works for negative numbers: = 4 = two s complement(4) = +! because starts with a and we are working with complements, we know it is a negative number. Which one? We complement it: s 2 s = - 2! Try this at home!! Gustavo Alonso, ETH Zürich. Programming in Assembly 3 The devil is in the details...! Modulus arithmetic and complements allows us to simplify arithmetic operations and cope with the limited space...! but we have to know how to interpret the result: "N (negative) bit = set if most significant bit is "Z (zero) bit = set if all bits are "V (overflow) bit, set if: in c = a - b, a and b have different signs and c and b have the same sign in c = a + b, a and b have the same sign but different from c! Conditions top use with signed arithmetic (two s complement)! compare two numbers is implemented as ion, e.g., compare -8 and = 6 = two s complement(6) = + Overflow (when overflow occurs, the resulting sign is the complement of the real sign) Gustavo Alonso, ETH Zürich. Programming in Assembly 4 Unsigned arithmetic! Unsigned arithmetic is, from the hardware point of view, identical to signed arithmetic! Overflow is now indicated by the C (carry) bit, which is set when there is a carry out of the most significant bit = 8 = + 6 = Carry set (ition)! Subtraction is also performed using two s complement, the C bit is set if there is no carry out of the most significant bit (C= indicates the result is negative) = 3 = two s complement(3)= + There is a carry, then C bit is not set (result is positive) = 2 = two s complements(2)= + There is no carry, then the C bit is set (result negative) Gustavo Alonso, ETH Zürich. Programming in Assembly 5 Branching with unsigned arithmetic Gustavo Alonso, ETH Zürich. Programming in Assembly 6

5 Multiplication (decimal) 23 x MULTIPLICAND MULTIPLIER Last digit of multiplier x multiplicand 2 x 23 = 46 Second to last digit of multiplier x multiplicand, shifted to the left one position 3 x 23 = 69 shift left() = 69 x = 69 Add them together Multiplication (decimal) with registers shift right shift right Gustavo Alonso, ETH Zürich. Programming in Assembly 7 Gustavo Alonso, ETH Zürich. Programming in Assembly 8 Multiplication (binary ) [I] Multiplication (binary ) [2] 3 = 5 = 3 = 5 = shift right shift right Gustavo Alonso, ETH Zürich. Programming in Assembly 9 Gustavo Alonso, ETH Zürich. Programming in Assembly 2

6 Multiplication (binary ) [3] Multiplication (binary ) [4] 3 = 5 = 3 = 5 = shift right shift right Gustavo Alonso, ETH Zürich. Programming in Assembly 2 Gustavo Alonso, ETH Zürich. Programming in Assembly 22 Multiplication (decimal, signed) 23 x -32 s complement(32)= = x 68 = 564 but the result should be -736 or, in complement form, What happened? 23 x -32 = 23 x ( - 32) = 23 x - 23 x 32 the result is too big by 23 x The problem is that the complement should be calculated based on the size of the result, not of the multiplier. Using the register multiplication algorithm, this only happens when the multiplier is negative 23 x -32 s complement(32) = = x 9968 = modulus = 9264 which is what we wanted. An alternative to get to the same result is to the excess (23 x ) (23 x ) s complement(23) = = 9264 again the result we wanted. This second solution has the advantage of requiring registers of smaller size. Multiplication signed (binary ) [I] shift right 3x-5 3 = -5 = Gustavo Alonso, ETH Zürich. Programming in Assembly 23 Gustavo Alonso, ETH Zürich. Programming in Assembly 24

7 Multiplication signed (binary ) [2] Multiplication (binary ) [3] 3 = -5 = 3 = -5 = shift right shift right Gustavo Alonso, ETH Zürich. Programming in Assembly 25 Gustavo Alonso, ETH Zürich. Programming in Assembly 26 Multiplication (binary ) [4] shift right 3 = -5 = two s complement Division (decimal) DIVISION IMPLEMENTED AS SUBTRACTION 737 / 32 D = 737 y. partial = ; d = 32 x /* y such that d equals D in digits */ 2. X = D 3. while ((X -= d) > ) y result += partial += d y 4. result -= ; partial -= d /* The loop executed one too many times */ 5. D = D - partial 6. if (D > 32) go to at the end, result contains the division and D the remainder Gustavo Alonso, ETH Zürich. Programming in Assembly 27 Gustavo Alonso, ETH Zürich. Programming in Assembly 28

8 Division (binary) with registers 737/ = 32 = two s complement(32) = RESULT shift right NEGATIVE! (went too far) shift right shift right shift right REMAINDER Gustavo Alonso, ETH Zürich. Programming in Assembly 29 Non-restoring division! Every time we go to far ing, we need to the divisor back, shift it and again. This can be simplified as follows.! Let a be the dividend and b the divisor. When the result of ing is negative, the next step is: a - b + b - b/2! working on that expression we get a -b + (2b - b)/2 a - b + b/2! that is, when we get a negative result, we do not need to the divisor back. It is enough to shift the divisor and it instead of ing Gustavo Alonso, ETH Zürich. Programming in Assembly 3 Non-restoring division shift right NEGATIVE! (went too far) shift right 737/ = 32 = two s complement(32) = shift right shift right RESULT REMAINDER Gustavo Alonso, ETH Zürich. Programming in Assembly 3