为什么我不能以这种方式释放分配的内存？

作为一项任务,我必须将两个字符串连接在一起并分配内存.完成后我希望能够free(*gluedstring)释放分配的内存.但现在我无法管理如何解决这个问题.

int strlen(char *s);
char *create_concatenated_cstring(char *source1, char *source2);
void destroy_cstring(char **gluedstring);

int main()
{
     char *string1 = "Common sense is genius ";
     char *string2 = "dressed in its working clothes.";
     char *together = create_concatenated_cstring(string1, string2);
     printf("Aan elkaar geplakt vormen de strings de volgende " \
     "quote:\n\n\"%s\"\n\n", together);
     destroy_cstring(&together);
     if (NULL == together)
        printf("De string was inderdaad vernietigd!\n");
     return 0;
}

int strlen(char *s)
{
     int n = 0;
     for (n = 0; *s != '\0'; s++, n++);
     return n;
}

char *create_concatenated_cstring(char *source1, char *source2)
{
    int size = strlen(source1) + strlen(source2) + 1;
    char *source3 = (char *)malloc(sizeof(char) * size);
    if(source3 == NULL)
    {
         printf("ERROR\n");
         return 0;
    }
    int i=0, j=0, k;
    int lengte1 = strlen(source1);
    int lengte2 = strlen(source2);
    for(;i

    

  




  
  
  

    

      

        Iharob Al As..
         5
      
      
因为你正在释放堆栈地址.您需要取消引用指针,就像这样

free(*gluedstring);
//   ^ `free' the pointer not it's address (or the pointer to it)
*gluedstring = NULL; // Prevent double `free' for example


注意,free()没有指针NULL,这就是为什么传递指针地址是好的,因为你可以设置它NULL并避免有悬空指针.


1> Iharob Al As..：
因为你正在释放堆栈地址.您需要取消引用指针,就像这样

free(*gluedstring);
//   ^ `free' the pointer not it's address (or the pointer to it)
*gluedstring = NULL; // Prevent double `free' for example


注意,free()没有指针NULL,这就是为什么传递指针地址是好的,因为你可以设置它NULL并避免有悬空指针.