Question

可能重复：
Why does simple C code receive segmentation fault?

我有下面的代码将从字符串中删除尾随空格但我不知道进入这段代码，因为它给出了分段错误问题??

 void main(void);

char* rtrim(char*);

void main(void)
 { 
 char* trail_str = "This string has trailing spaces in it.               ";


 printf("Before calling rtrim(), trail_str is '%s'\n", trail_str);

 printf("and has a length of %d.\n", strlen(trail_str));



 rtrim(trail_str);



 printf("After calling rtrim(), trail_str is '%s'\n", trail_str);

 printf("and has a length of %d.\n", strlen(trail_str));

 }


  char* rtrim(char* str)
  {
   int n = strlen(str) - 1;    

    while (n>0)            
   {
       if (*(str+n) != ' ')    
      {

           *(str+n+1) = '\0'; 



           break;             
      }

      else

      n--;
 }

 return str;      

 }

Answer 1

trail_str指向内存中的常量区域，因此无法在*(str+n+1) = '\0'

中更改初始化时

char* trail_str = "This string has trailing spaces in it.               ";

你实际生成一个常量字符串："This string has trailing spaces in it. "并告诉trail_str指向它。

Answer 2

char* trail_str = "This string has trailing spaces in it.               ";

tail_str指向的字符串可以存储在只读内存中。你不能修改它。

如果要修改它，则需要为其分配存储空间并将该字符串复制为常量。

（另外，main应该返回int，而不是void。）

为什么下面的代码会出现分段错误问

2 个答案: