更改按值传递的指针 [英] change pointer passed by value

问题描述

我给了一个函数foo(struct node *n)，其中n是链表中的头节点. 现在foo应该更改ns.t.它指向列表的末尾.

I have given a function foo(struct node *n) where n is the head node in a linked list. Now foo should change n s.t. it points to the end of the list.

但是这个函数签名有可能吗?

But is this possible with this function signature?

假设t是指向列表末尾的指针:

Assuming t is the pointer to the end of the list:

• n = t将不起作用，因为指针是按值传递的.
• *n = *t不起作用，因为我会覆盖列表的开头.

• n = t won't work because the pointer is passed by value.
• *n = *t won't work because I would overwrite the head of the list.

我错过了什么吗?

推荐答案

您将需要使用指向指针的指针:

You would need to use a pointer to a pointer:

foo(struct node **n) 

要更改n指向的内容，请执行以下操作:

To change what n points to, you do:

*n = t;

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