一个处理字符串的问题?

[Sylix]

如:
char* p1 = "ewew";


怎么在这个字符串最前面添加字符串呢?

变成
char* p1 ="11ewew";

[lw1a2]

笨办法：
int i;
char p1[7]="ewew";
char p2[7]="ll";
strcat(p2,p1);
for(i=0;i<7;i++)
p1[i]=p2[i];

[Flood1984]

直接在这个前面加是不安全的，
因为他前面和后面也许根本没有空间来存放。

[winstonch]

只能用笨办法

[ra3]

char* p1 = "ewew";

p1 指向的是个常量, 所以不能添;

只有用 char p1[N] = "ewew";

[qingyuan18]

char *p1 = (char *)malloc(sizeof(10));
strcpy(p1,"11ewew");

[dongpy]

char* p1 = "ewew";是常量字符串，要把它存放到栈上或堆上才能添。

char s1[7];
strcpy(s1+2, "ewew");

char s2[3] = "ll";
memcpy(s1, s2, 2);

[x_wgc]

char *p1 = (char *)malloc(sizeof(10));
strcpy(p1,"11ewew");

同意这个

[delhe]

int len1,len2;
char* p1 = "ewew";
char* p2 = "ll";
len1 = length(*p1);
len2 = length(*p2);
char* p3;
p3 = (char*)malloc(sizeof(len1+len2+1));
while(*p3++ = *p2++);
while (*p3++ = *p1++) ;
*p3 = "\0";

[ycom__net]

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{
int len1, len2;
char *str1="werrt";
char *str2="sdfwerwr";
len1=strlen(str1);
len2=strlen(str2);
char *str3=(char*)malloc(sizeof(len1+len2+1));
strcpy(str3,str2);
strcpy(str3+len2, str1);
*(str3+len1+len2)='\0';
printf("%s",str3);
system("PAUSE");
return 0;
}