求字符串长度理论（不是简单的用循环的）

求字符串长度理论

不用判断语句，不用循环语句，只有顺序，如何求得字符串长度，请各位赐教~！！！！

[解决办法]
C:\Program Files\Microsoft Visual Studio 10.0\VC\crt\src\intel\strlen.asm

Assembly code

        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        .listpage;***;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  strlenstrlen  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_loopstr_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 redundantmain_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 setbyte_3:        lea     eax,[ecx - 1]        mov     ecx,string        sub     eax,ecx        retbyte_2:        lea     eax,[ecx - 2]        mov     ecx,string        sub     eax,ecx        retbyte_1:        lea     eax,[ecx - 3]        mov     ecx,string        sub     eax,ecx        retbyte_0:        lea     eax,[ecx - 4]        mov     ecx,string        sub     eax,ecx        retstrlen  endp        end