C语言:释放指向struct的指针的内存

问题描述

说我已经声明了一个指向结构的指针，并使用此定义将其分配给malloc()

Say I have declared a pointer to a struct and assign it with malloc() using this definition

typedef struct node {
    int info;
    struct node *next;
} NODE;

然后在代码中的某个地方我声明了两个指向它的指针

Then somewhere in the code I declared two pointers to it

NODE *node1, *node2 = NULL;

node1 = malloc(sizeof(NODE));
node2 = node1;

我的问题是，是否应该像人们总是通过 free(node1)对node1一样使用"free()"来释放node2.分配 node2 = node1;

My question, should I use "free()" to release node2 just like people always do to node1 via free(node1). What's exactly the effect of the assignment node2 = node1;

谢谢.

推荐答案

执行时

node1 = malloc(sizeof(NODE));

你有类似的东西

+-------+      +-----------------------------+
| node1 | ---> | memory for a NODE structure |
+-------+      +-----------------------------+

在分配了node2 = node1之后，您有了:

After the assignment node2 = node1 you have instead this:

+-------+
| node1 | -\
+-------+   \    +-----------------------------+
             >-> | memory for a NODE structure |
+-------+   /    +-----------------------------+
| node2 | -/
+-------+

换句话说，您有两个指针指向相同内存.

In other words you have two pointers pointing to the same memory.

尝试使用两个指针变量之一调用free会使两个指针无效.

Attempting to call free using either of the two pointer variable will invalidate both pointers.

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