问个memcmp的问题，想修改他最后的返回值，有点牵涉到汇编的，懂的进来看看。

[jpsr]

请问memcmp这个函数他返回的效果和strcmp一样的，如果我想要memcmp来返回2个连续内存中首个不相同的地址，那么应该怎么去修改这个memcmp？？全是汇编啊，看不懂 @@
有会的请帮我看看怎么修改？？我个人觉得汇编可能比较快一点，所以想改这段汇编代码，请看清问题再回答。
附memcmp的汇编代码：
CODESEG

public memcmp
memcmp proc

.FPO ( 0, 3, 0, 0, 0, 0 )

mov eax,[esp+0ch] ; eax = counter
test eax,eax ; test if counter is zero
jz short retnull ; return 0

mov edx,[esp+4] ; edx = buf1
push esi
push edi
mov esi,edx ; esi = buf1
mov edi,[esp+10h] ; edi = buf2

; Check for dword (32 bit) alignment
or edx,edi
and edx,3 ; edx=0 iff buf1 are buf2 are aligned
jz short dwords

; Strings are not aligned. If the caller knows the strings (buf1 and buf2) are
; different, the function may be called with length like -1. The difference
; may be found in the last dword of aligned string, and because the other
; string is misaligned it may cause page fault. So, to be safe. the comparison
; must be done byte by byte.
test eax,1
jz short main_loop

mov cl,[esi]
cmp cl,[edi]
jne short not_equal
inc esi
inc edi
dec eax
jz short done ; eax is already 0

main_loop:
mov cl,[esi]
mov dl,[edi]
cmp cl,dl
jne short not_equal

mov cl,[esi+1]
mov dl,[edi+1]
cmp cl,dl
jne short not_equal

add edi,2
add esi,2

sub eax,2
jnz short main_loop
done:
pop edi
pop esi
retnull:
ret ; _cdecl return


dwords:
mov ecx,eax
and eax,3 ; eax= counter for tail loop

shr ecx,2
jz short tail_loop_start
; counter was >=4 so may check one dword
rep cmpsd

jz short tail_loop_start

; in last dword was difference
mov ecx,[esi-4] ; load last dword from buf1 to ecx
mov edx,[edi-4] ; load last dword from buf2 to edx
cmp cl,dl ; test first bytes
jne short difference_in_tail
cmp ch,dh ; test seconds bytes
jne short difference_in_tail
shr ecx,10h
shr edx,10h
cmp cl,dl ; test third bytes
jne short difference_in_tail
cmp ch,dh ; they are different, but each one is bigger?
; jmp short difference_in_tail

difference_in_tail:
mov eax,0
; buf1 < buf2 buf1 > buf2
not_equal:
sbb eax,eax ; AX=-1, CY=1 AX=0, CY=0
pop edi ; counter
sbb eax,-1 ; AX=-1 AX=1
pop esi
ret ; _cdecl return

; in tail loop we test last three bytes (esi and edi are aligned on dword
; boundary)
tail_loop_start:

test eax,eax ; eax is counter%4 (number of bytes for tail
; loop)
jz short done ; taken if there is no tail bytes
mov edx,[esi] ; load dword from buf1
mov ecx,[edi] ; load dword from buf2
cmp dl,cl ; test first bytes
jne short difference_in_tail
dec eax ; counter--
jz short tail_done
cmp dh,ch ; test second bytes
jne short difference_in_tail
dec eax ; counter--
jz short tail_done
and ecx,00ff0000h ; test third bytes
and edx,00ff0000h
cmp edx,ecx
jne short difference_in_tail
dec eax
tail_done:
pop edi
pop esi
ret ; _cdecl return

memcmp endp
end

[Wenxy1]

晕倒，Win 32 的偶不熟，帮你UP

[try876]

UP

[jpsr]

谢谢，我自己up

[jpsr]

高手都不愿意回答，我们自己学习asm，问题已经自己解决了，有会的，请贴一下你修改后的memcmp，好让我结帐。

[riverboat]

自己用字符串操作写一个吧，应该也不太难啊。

size_t memcmp1(const void *p1, const void * p2, size_t size){
size_t i;
void *q1=p1, *q2=p2;

for(i=0; i<size; i++, q1++, q2++){
if((*q1)!=(*q2)) return i; // 一个个字节比较，不一样就返回该位置
}
return size; // 所有字节都一样，表明相等
}

[jpsr]

我要用rep cmpsb，不用for语句。我听别人说用cmpsd比较快，不过这里有个对齐的问题，还没搞明白。

[双杯献酒]

我的建议是不要修改它。

[jpsr]

我把memcmp的代码搬到程序里面了，memcmp.asm我没修改
就是
__asm
{
……
}