c - 使链表的节点成为最后一个节点的功能？

Question

此函数将获取链表的特定节点并将其作为最后节点。

void lastNode(struct list *node)
{
  struct list* temp = node->next;
  node->next = NULL;
  struct list* temp2;
  while(temp)
    {

      temp2 = temp->next;
      printf("%p \n",temp);
      free(temp);
      temp = temp2;
    }

}

int main()
{
  struct list d={4,NULL};
  struct list c={3,&d};
  struct list b={2,&c};
  struct list a={1,&b};


  struct list *node = &a;

  lastNode(&b);

  while(node)
    {
      printf("%d \n",node->d);
      node=node->next;
    }
  return 0;
}

但在释放节点时出现错误。

Answer 1

查看您的完整代码:

struct list c={3,&d};
struct list b={2,&c};

lastNode(&b);

free temp 是不合法的，因为它不是使用 malloc 分配的。

struct list* temp = node->next; /* temp is now &c. You can't free it. */
while(temp)
{
    /* ... */  
    free(temp); /* Tries to free &c which is on the stack. */
}