The greatest common divisor (GCD) of two integers is the largest integer that wi

ID: 3775510 • Letter: T

Question

The greatest common divisor (GCD) of two integers is the largest integer that will evenly divide
both integers. The GCD algorithm involves integer division in a loop, described by the following
pseudocode:
int GCD(int x, int y)
{
     x = abs(x)   // absolute value
     y = abs(y)
     do {
               int n = x % y
               x = y
               y = n
               } while (y > 0)
     return x
}
Implement this function in assembly language and write a test program that calls the function
several times, passing it different values. Display all results on the screen.

I have this codes but it doesn't work.

INCLUDE Irvine32.inc

.data
array SDWORD -5,-20,18,24,11,7,438,226,13,-26
str1 BYTE "Greatest common divisor is: ",0

.code
main PROC

mov ecx,LENGTHOF array / 2
mov esi,OFFSET array

L1:   mov eax,[esi]
   mov ebx,[esi+4]
   call CalcGcd
   mov edx,OFFSET str1
   call WriteString
   call WriteDec
   call Crlf
   add esi,TYPE array * 2
   loop L1

exit
main ENDP

CalcGcd PROC

push ebx
push edx

   .IF SDWORD PTR eax < 0
   neg eax
   .ENDIF

   .IF SDWORD PTR ebx < 0
   neg ebx
   .ENDIF

L1:   mov edx,0
   div ebx
   cmp edx,0
   je L2   ; yes: quit
   mov eax,ebx
   mov ebx,edx
   jmp L1

L2: mov eax,ebx

   pop edx
   pop ebx
   ret
CalcGcd ENDP

END main

Explanation / Answer

hope this is usefull

INCLUDE Irvine32.inc
.data
array SDWORD -5,-20,18,24,11,7,438,226,13,-26
str1 BYTE "Greatest common divisor is: ",0
.code
main PROC
mov ecx,LENGTHOF array / 2
mov esi,OFFSET array
L1: mov eax,[esi]
mov ebx,[esi+4]
call CalcGcd
mov edx,OFFSET str1
call WriteString
call WriteDec
call Crlf
add esi,TYPE array * 2
loop L1
exit
main ENDP

push ebx
push edx
.IF SDWORD PTR eax < 0 ; force a signed comparison
neg eax ; EAX = abs(EAX)
.ENDIF
.IF SDWORD PTR ebx < 0
neg ebx ; EBX = abs(EBX)
.ENDIF
L1: mov edx,0
div ebx ; divide int1 by int2
cmp edx,0 ; remainder = 0?
je L2 ; yes: quit
mov eax,ebx ; no: prepare for
mov ebx,edx ; next iteration
jmp L1
L2: mov eax,ebx ; EAX = GCD
pop edx
pop ebx
ret
CalcGcd ENDP
END main

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