c - 如何指向一个对象而不在删除时影响它？

Question

struct object
{
   char* whatever;
}
struct other_object
{
   object** points;
   int number;
}

void add_to_points(other_object* passer, const struct object* object_ptr)
{
   passer->points[passer->number] = object_ptr; //leads to warning
   number++;
}

void delete_memory(other_object* passer)
{
   int i;
   for(i = 0; i < passer->numbers; i++){
      free(passer->points[i]);
   }
}

struct object* new_object = malloc(sizeof(object));
struct other_object* new_object_2 = malloc(sizeof(other_object));
new_object_2->points = malloc(3*sizeof(object));

add_to_points(new_object_2, new_object);
delete_memory(new_object_2);

所以通过上面的代码，看起来delete_memory实际上是在释放对象本身。我知道这可能与我的 add_to_points 函数有关，但我不清楚如何更改它以便“点”仅指向对象，但是如果我们删除点的成员(例如点[2])或者整个物体在一起，物体本身并没有被破坏。

Answer 1

您必须使用 object* 而不是 object 调用 malloc 来获取 new_object_2->points。您似乎想要获得指向对象的指针数组，而不是像您那样获得对象数组。

替换

new_object_2->points = malloc(3*sizeof(object));

与

new_object_2->points = malloc(3*sizeof(object*));