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回复 #1 Solidus 的帖子
代码有问题
这个差不多
size_t strlen(const char *_string)
{
const char *str = _string;
while ( *str && ((DWORD)str & (sizeof(DWORD) - 1)) ) str++;
if ( ! *str ) return str - _string;
DWORD magic_bits = 0x7EFEFEFF;
const DWORD *_ptr = (DWORD *)str;
DWORD tmp_data;
do {
tmp_data = *_ptr++;
} while ( (~tmp_data ^ (tmp_data + magic_bits)) == ~magic_bits);
for ( str = (const char *)(_ptr - 1); *str; str++ );
return str - _string;
} |
微软的汇编代码
page ,132
title strlen - return the length of a null-terminated string
;***
;strlen.asm - contains strlen() routine
;
; Copyright (c) Microsoft Corporation. All rights reserved.
;
;Purpose:
; strlen returns the length of a null-terminated string,
; not including the null byte itself.
;
;*******************************************************************************
.xlist
include cruntime.inc
.list
page
;***
;strlen - return the length of a null-terminated string
;
;Purpose:
; Finds the length in bytes of the given string, not including
; the final null character.
;
; Algorithm:
; int strlen (const char * str)
; {
; int length = 0;
;
; while( *str++ )
; ++length;
;
; return( length );
; }
;
;Entry:
; const char * str - string whose length is to be computed
;
;Exit:
; EAX = length of the string "str", exclusive of the final null byte
;
;Uses:
; EAX, ECX, EDX
;
;Exceptions:
;
;*******************************************************************************
CODESEG
public strlen
strlen proc \
buf:ptr byte
OPTION PROLOGUE:NONE, EPILOGUE:NONE
.FPO ( 0, 1, 0, 0, 0, 0 )
string equ [esp + 4]
mov ecx,string ; ecx -> string
test ecx,3 ; test if string is aligned on 32 bits
je short main_loop
str_misaligned:
; simple byte loop until string is aligned
mov al,byte ptr [ecx]
add ecx,1
test al,al
je short byte_3
test ecx,3
jne short str_misaligned
add eax,dword ptr 0 ; 5 byte nop to align label below
align 16 ; should be redundant
main_loop:
mov eax,dword ptr [ecx] ; read 4 bytes
mov edx,7efefeffh
add edx,eax
xor eax,-1
xor eax,edx
add ecx,4
test eax,81010100h
je short main_loop
; found zero byte in the loop
mov eax,[ecx - 4]
test al,al ; is it byte 0
je short byte_0
test ah,ah ; is it byte 1
je short byte_1
test eax,00ff0000h ; is it byte 2
je short byte_2
test eax,0ff000000h ; is it byte 3
je short byte_3
jmp short main_loop ; taken if bits 24-30 are clear and bit
; 31 is set
byte_3:
lea eax,[ecx - 1]
mov ecx,string
sub eax,ecx
ret
byte_2:
lea eax,[ecx - 2]
mov ecx,string
sub eax,ecx
ret
byte_1:
lea eax,[ecx - 3]
mov ecx,string
sub eax,ecx
ret
byte_0:
lea eax,[ecx - 4]
mov ecx,string
sub eax,ecx
ret
strlen endp
end
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发表于 2009-07-09 15:36
发表于 2009-07-09 16:28
