请问：字符串中字符倒序排列，不借用另一字符串

请教：字符串中字符倒序排列，不借用另一字符串
请教各位达人，如何将字符串中的字符在同一字符串中倒序排列，不借用另一字符串。例如，输入串为“abcd”，则将该串变为”dcba“。这是含有偶数个字符的串，如果是含有奇数个字符的串，可能就又又同了。我做了一个函数，但达不到效果。

void reverse(char *str)
{
int i, j;
char temp;

for(i = 0, j = strlen(str)-1;
i <= strlen(str)/2, j > strlen(str)/2;
i++, j--)
{
temp = str[i];
str[i] = str[j];
str[j] = temp;
}
}

麻烦帮忙看看问题出在哪里，谢谢！
[解决办法]
试试这个：
j = strlen(str);
for(i = 0; i <= j / 2; i++)
{
temp = str[i];
str[i] = str[j - i];
str[j - i] = temp;
}
[解决办法]
i <= strlen(str)/2, j > strlen(str)/2
改成
(i <= strlen(str)/2) && (j > strlen(str)/2)
[解决办法]

版本1
void reverse(char *str)
{
int i;
int j = strlen(str);
char temp;
for (i=0; i&lt;j;i++,j--)
{
temp = str[i];
str[i] = str[j-1];
str[j-1] = temp;
}
}

版本2
void reverse(char *str)
{
int i;
int j = strlen(str);
char temp;
for (i=0; i&lt;j/2;i++)
{
temp = str[i];
str[i] = str[j-i-1];
str[j-i-1] = temp;
}
}

版本3
void reverse(char *str)
{
char temp;
char *last_char;

for (last_char = str; *last_char != '\0'; last_char++)
;
last_char--;

while (str &lt; last_char)
{
temp = *str;
*str++ = *last_char;
*last_char-- = temp;
}
}

版本4
void reverse(char *str)
{
char temp;
char *last_char;

for (last_char = str; *last_char != '\0'; last_char++)
;
last_char--;
while(str&lt;last_char)
{
*str = *str ^ *last_char;
*last_char = *str ^ *last_char;
*str = *str ^ *last_char;;
str++;
last_char--;
}
}

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

        page    ,132
        title   strrev - reverse a string in place
;***
;strrev.asm - reverse a string in place
;
;       Copyright (c) Microsoft Corporation. All rights reserved.
;
;Purpose:
;       defines _strrev() - reverse a string in place (not including
;       '\0' character)
;
;*******************************************************************************

        .xlist
        include cruntime.inc
        .list

page
;***
;char *_strrev(string) - reverse a string in place
;
;Purpose:
;       Reverses the order of characters in the string.  The terminating
;       null character remains in place.
;
;       Algorithm:
;       char *
;       _strrev (string)
;             char *string;
;             {
;             char *start = string;
;             char *left = string;
;             char ch;
;
;             while (*string++)
;                     ; 
 
;             string -= 2;
;             while (left &lt; string)
;                     {
;                     ch = *left;
;                     *left++ = *string;
;                     *string-- = ch;
;                     }
;             return(start);
;             }
;
;       NOTE: There is a check for an empty string in the following code.
;       Normally, this would fall out of the "cmp si,di" instruction in the
;       loop portion of the routine.  However, if the offset of the empty
;       string is 0 (as it could be in large model), then the cmp does not
;       catch the empty string and the routine essentially hangs (i.e., loops
;       moving bytes one at a time FFFFh times).  An explicit empty string
;       check corrects this.
;
;Entry:
;       char *string - string to reverse
;
;Exit:
;       returns string - now with reversed characters
;
;Uses:
;
;Exceptions:
;
;*******************************************************************************

        CODESEG

        public  _strrev
_strrev proc \
        uses edi esi, \
        string:ptr byte

        mov     edi,[string]    ; di = string
        mov     edx,edi         ; dx=pointer to string; save return value

        mov     esi,edi         ; si=pointer to string
        xor     eax,eax         ; search value (null)
        or      ecx,-1          ; cx = -1
repne   scasb                   ; find null
        cmp     ecx,-2          ; is string empty? (if offset value is 0, the
        je      short done      ; cmp below will not catch it and we'll hang).

        sub     edi,2           ; string is not empty, move di pointer back
                                ; di points to last non-null byte

lupe:
        cmp     esi,edi         ; see if pointers have crossed yet
        jae     short done      ; exit when pointers meet (or cross)

        mov     ah,[esi]        ; get front byte...
        mov     al,[edi]        ;   and end byte
        mov     [esi],al        ; put end byte in front... 
 
        mov     [edi],ah        ;   and front byte at end
        add     esi,1           ; front moves up...
        sub     edi,1           ;   and end moves down
        jmp     short lupe      ; keep switching bytes

done:
        mov     eax,edx         ; return value: string addr

        ret                     ; _cdecl return

_strrev endp
        end

[解决办法]
C:\Program Files\Microsoft Visual Studio 10.0\VC\crt\src\strrev.c

/***
*strrev.c - reverse a string in place
*
*       Copyright (c) Microsoft Corporation. All rights reserved.
*
*Purpose:
*       defines _strrev() - reverse a string in place (not including
*       '\0' character)
*
*******************************************************************************/

#include &lt;cruntime.h&gt;
#include &lt;string.h&gt;

/***
*char *_strrev(string) - reverse a string in place
*
*Purpose:
*       Reverses the order of characters in the string.  The terminating
*       null character remains in place.
*
*Entry:
*       char *string - string to reverse
*
*Exit:
*       returns string - now with reversed characters
*
*Exceptions:
*
*******************************************************************************/

char * __cdecl _strrev (
        char * string
        )
{
        char *start = string;
        char *left = string;
        char ch;

        while (*string++)                 /* find end of string */
                ;
        string -= 2;

        while (left &lt; string)
        {
                ch = *left;
                *left++ = *string;
                *string-- = ch;
        }

        return(start);
}

请问：字符串中字符倒序排列不借用另

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