如何在C中正确释放char **表 [英] how to properly free a char **table in C

问题描述

我需要您对这段代码的建议: 表字段options [0]，options [1]等...似乎没有正确释放. 感谢您的回答

I need your advice on this piece of code: the table fields options[0], options[1] etc... doesn't seem to be freed correctly. Thanks for your answers

int main()
{
  ....
  char **options;
  options = generate_fields(user_input);
  for(i = 0; i < sizeof(options) / sizeof(options[0]); i++)  {
    free(options[i]);
    options[i] = NULL;
  }

  free(options);
}

char ** generate_fields(char *) 
{
   char ** options = malloc(256*sizeof(char *));
   ...
   return options;

}

推荐答案

问题是这样的:

for(i = 0; i < sizeof(options) / sizeof(options[0]); i++)

options是指针类型，而不是数组类型，因此sizeof(options)将始终相同(在32位计算机上通常为4个字节，在64位计算机上通常为8个字节)，因此sizeof(options)/sizeof(options[0])几乎总是1.

options is a pointer type, not an array type, so sizeof(options) will always be the same (typically 4 bytes on a 32-bit machine or 8 bytes on a 64-bit machine), so sizeof(options)/sizeof(options[0]) will almost always be 1.

关键是要始终按与malloc相同的方式存储free内存.因此，如果先malloc二维数组，然后malloc一系列一维数组，则在释放它时需要做相反的操作:

The key is to always free memory in the same manner as you malloc'ed it. So, if you malloc a 2-dimensional array and then malloc a series of 1-dimensional arrays, you need to do the reverse when freeing it:

char ** generate_fields(char *) 
{
   char ** options = malloc(256*sizeof(char *));
   for(int i = 0; i < 256; i++)
       options[i] = malloc(some_size);
   return options;
}

void free_fields(char ** options)
{
    for(int i = 0; i < 256; i++)
        free(options[i]);
    free(options);
}

请注意，如果大小(在这种情况下为256)不是常数，则您需要自己对其进行跟踪，因为否则，您将无法知道释放时要循环多少次.

Note that if the size (256 in this case) is not a constant, you need to keep track of it yourself, since otherwise you have no way of knowing how many times to loop when freeing.

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