C语言strcmp函数是什么样的代码?

Question

C语言strcmp函数是什么样的代码

Answer 1

strcmp（s1，s2）
相同返回0
当s1<s2时，返回值<0 当s1=s2时，返回值=0 当s1>s2时，返回值>0 即：两个字符串自左向右逐个字符相比（按ASCII值大小相比较），直到出现不同的字符或遇'\0'为止。如： "A"<"B" "a">"A" "computer">"compare" 特别注意：strcmp(s1,s2)这里面只能比较字符串，不能比较数字等其他形式的参数。

Answer 2

page ,132
title strcmp.asm - compare two strings
;***
;strcmp.asm - routine to compare two strings (for equal, less, or greater)
;
; Copyright (c) Microsoft Corporation. All rights reserved.
;
;Purpose:
; STRCMP compares two strings and returns an integer
; to indicate whether the first is less than the second, the two are
; equal, or whether the first is greater than the second, respectively.
; Comparison is done byte by byte on an UNSIGNED basis, which is to
; say that Null (0) is less than any other character (1-255).
;
;*******************************************************************************

.xlist
include cruntime.inc
.list

page
;***
;strcmp - compare two strings, returning less than, equal to, or greater than
;
;Purpose:
; Compares two string, determining their lexical order. Unsigned
; comparison is used.
;
; Algorithm:
; int strcmp ( src , dst )
; unsigned char *src;
; unsigned char *dst;
; {
; int ret = 0 ;
;
; while( ! (ret = *src - *dst) && *dst)
; ++src, ++dst;
;
; if ( ret < 0 )
; ret = -1 ;
; else if ( ret > 0 )
; ret = 1 ;
;
; return( ret );
; }
;
;Entry:
; const char * src - string for left-hand side of comparison
; const char * dst - string for right-hand side of comparison
;
;Exit:
; AX < 0, 0, or >0, indicating whether the first string is
; Less than, Equal to, or Greater than the second string.
;
;Uses:
; CX, DX
;
;Exceptions:
;
;*******************************************************************************

CODESEG

public strcmp
strcmp proc \
str1:ptr byte, \
str2:ptr byte

OPTION PROLOGUE:NONE, EPILOGUE:NONE

.FPO ( 0, 2, 0, 0, 0, 0 )

mov edx,[esp + 4] ; edx = src
mov ecx,[esp + 8] ; ecx = dst

test edx,3
jnz short dopartial

align 4
dodwords:
mov eax,[edx]

cmp al,[ecx]
jne short donene
or al,al
jz short doneeq
cmp ah,[ecx + 1]
jne short donene
or ah,ah
jz short doneeq

shr eax,16

cmp al,[ecx + 2]
jne short donene
or al,al
jz short doneeq
cmp ah,[ecx + 3]
jne short donene
add ecx,4
add edx,4
or ah,ah
jnz short dodwords

align 4
doneeq:
xor eax,eax
ret

align 4
donene:
; The instructions below should place -1 in eax if src < dst,
; and 1 in eax if src > dst.

sbb eax,eax
sal eax,1
add eax,1
ret

align 4
dopartial:
test edx,1
jz short doword

mov al,[edx]
add edx,1
cmp al,[ecx]
jne short donene
add ecx,1
or al,al
jz short doneeq

test edx,2
jz short dodwords

align 4
doword:
mov ax,[edx]
add edx,2
cmp al,[ecx]
jne short donene
or al,al
jz short doneeq
cmp ah,[ecx + 1]
jne short donene
or ah,ah
jz short doneeq
add ecx,2
jmp short dodwords

strcmp endp