Cleanup time....dun dun DUN!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

                                      On this post I'm going to show you how I cleaned up my previous program and explain why.


Here's my modified program:




.586
.model flat, c
.stack 100h
.data

; gcd(10, 4), gcd(31415, 14142), gcd(3241343, 948324),
; gcd(31243,68455), gcd(75324,729345), gcd(327256,647384)
left dword 10, 31415, 3241343, 31243, 75324, 327256
right dword 4, 14142, 948324, 68455, 729345, 647384


.code

doit proc
  xor ecx,ecx
doagain:
    push left[ecx]
push right[ecx]
call gcd
add esp,8
add ecx,4

cmp ecx,24
jle doagain
 ret

doit endp

gcd proc
mov eax,[esp + 8]       ; eax is left
mov ebx,[esp + 4]       ; ebx is right
again:
xor edx,edx                 ; zeroing out edx to prevent Integer overflow
div ebx                        ; divides left by right
cmp edx, 0     ; We are done when remainder is zero
je weAreDone

mov eax,ebx     ; we have a new left
mov ebx,edx               ; our remainder is our new right
jmp again                    ; repeat

weAreDone:
; ebx has the greatest common divisor
    ret

gcd endp

end

                            First I moved my compare from the end of gcd into the middle,and the reason we do that is because if we leave the compare at the end then once we've got our greatest common divisor then every thing else after that would just be extra work which is why I moved the compare to right after the division problem,if the 2nd number is 0 we jump straight to the return,and if it's not we just continue and switch all the numbers. Also I changed it so that we only read off of the inputs and we don't change them at all,because it's like doit sent a letter to gcd and if we change the inputs it's like we drew a picture all over that letter. And now since we use the inputs a lot less we use the registers a lot more,and a good thing about that is using the registers is way faster. Also I changed my commentary.

WELL TTFN,TA TA FOR NOW!!!!!!!!!!!!!!!!!!!!