为什么在字符指针上使用 strcat 会崩溃? [英] Why does using strcat on character pointers crash?

问题描述

为什么这段代码会崩溃?在字符指针上使用 strcat 是非法的吗?

why does this code crash? is using strcat illegal on character pointers?

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

int main()
{
   char *s1 = "Hello, ";
   char *s2 = "world!";
   char *s3 = strcat(s1, s2);
   printf("%s",s3);
   return 0;
}

请给出引用数组和指针的正确方法.

please give a proper way with referring to both array and pointers.

推荐答案

问题是 s1 指向一个字符串文字，而您试图通过附加 s2 给它.不允许修改字符串文字.您需要创建一个字符数组并将两个字符串复制到其中，如下所示:

The problem is that s1 points to a string literal and you are trying to modify it by appending s2 to it. You are not allowed to modify string literals. You need to create a character array and copy both strings into it, like so:

char *s1 = "Hello, ";
char *s2 = "world!";

char s3[100] = ""; /* note that it must be large enough! */
strcat(s3, s1);
strcat(s3, s2);
printf("%s", s3);

足够大"意味着至少 strlen(s1) + strlen(s2) + 1.+ 1 是为了说明空终止符.

"Large enough" means at least strlen(s1) + strlen(s2) + 1. The + 1 is to account for the null terminator.

话虽如此，您应该认真考虑使用经过边界检查的 strncat(或者可以说更好但非标准的 strlcat，如果可用)，并且因此远远优于 strcat.

That having been said, you should seriously consider using strncat (or the arguably better but nonstandard strlcat, if it is available), which are bounds-checked, and thus are far superior to strcat.

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