c - 使用函数调用中函数返回的指针

Question

Is the following usage of pointers in functions call is a memory leak:

bson_t * parse_json(const char * json_fields){

    bson_error_t error;
    bson_t *bson_fields = bson_new_from_json((unsigned char *)json_fields, -1, &error);
    if (!bson_fields) {
        log_die("Error: %s\n", error.message);
    } else {
      return bson_fields;
    }
    log_die("Error: something bad happend in parse_columns");
    return bson_fields; // this should never be reached ...
}

The following code works, but what happens to the pointer from parse_json here? Is this a memory leak?

bson_concat(fields, parse_json(json_fields));

The mongodb C-API offers the function bson_destory :

bson_destroy(fields);

I am wondering maybe it's better to explicitly free the memory of new_fields :

        bson_t *new_fields = parse_json(json_fields);
        bson_concat(fields, new_fields);
        bson_destroy(new_fields);

While this example uses mongodb c-api, I am also trying to understand the general case.

  some_type * pointer_returner(){
  some_type *var;
  ...

  return var;
  }


  do_something(pointer_retuner());

Is the call above causing a memory leak?

Answer 1

Yes, you need to call bson_destroy to deallocate your structure object it is no longer used.

From bson_destroy documentation :

The bson_destroy() function shall free an allocated bson_t structure.

This function should always be called when you are done with a bson_t unless otherwise specified.