怎么用递归方法将一个字符串倒序

[mo01]

函数原型为void reserve(char* s);

[xghome]

void reserve(char* s)
{
static char s1[256];
static int i=0;
static int j=255;
if('\0'!=s[i])
{
s1[j]=s[i];
i++;
j--;
reserve(s);
}
}

[lkcowboy]

动态数组更好

[kmzwm]

void reserve(char* s)
{
char c;
if (s == NULL || *s == '\0')
return;
if(*s != 0)
{

*s = *(s+ strlen(s) -1);

reserve(s+1);
}

}

[kmzwm]

void reserve(char* s)
{
char beg;
char *pStr1;
char *pStr2 ;
if (s == NULL || *s == '\0')
return;
//move right to allocate the last char
pStr1 = pStr2 = s + strlen(s) - 2;
while ( pStr1 <= s+1 )
{
pStr1 --;
*pStr2-- = *pStr1;
}

*s = *(s+ strlen(s) -1);
reserve(s+1);

}

[mo01]

我刚才试过了，两位的方法都是对的，谢谢了！

[kuhx]

我也来一个：
void reserve(char* s)
{
int i = strlen( s );
char* newchar = new char[i];
ZeroMemory(newchar , i);
for ( int j = 0 ; j < i ; j++ )
{
newchar[j] = s[i-j-1];
}
ZeroMemory(s , i);
for ( int j = 0 ; j < i ; j++ )
{
s[j] = newchar[i-j-1];
}
}