为什么链表使用指针而不是将节点存储在节点内 [英] Why do linked lists use pointers instead of storing nodes inside of nodes

问题描述

我之前在 Java 中广泛使用过链表，但我对 C++ 非常陌生.我正在使用在项目中提供给我的这个节点类很好

I've worked with linked lists before extensively in Java, but I'm very new to C++. I was using this node class that was given to me in a project just fine

class Node
{
  public:
   Node(int data);

   int m_data;
   Node *m_next;
};

但是我有一个问题没有得到很好的回答.为什么要使用

but I had one question that wasn't answered very well. Why is it necessary to use

Node *m_next;

指向列表中的下一个节点而不是

to point to the next node in the list instead of

Node m_next;

我理解最好使用指针版本；我不会争论事实，但我不知道为什么这样更好.关于指针如何更好地进行内存分配，我得到了一个不太明确的答案，我想知道这里是否有人可以帮助我更好地理解这一点.

I understand that it is better to use the pointer version; I'm not going to argue facts, but I don't know why it's better. I got a not so clear answer about how the pointer is better for memory allocation, and I was wondering if anyone here could help me understand that better.

推荐答案

这不仅更好，而且是唯一可能的方法.

It's not just better, it's the only possible way.

如果你在自身内部存储了一个 Node object，那么 sizeof(Node) 会是什么?它将是 sizeof(int) + sizeof(Node)，这将等于 sizeof(int) + (sizeof(int) + sizeof(Node))，即将等于 sizeof(int) + (sizeof(int) + (sizeof(int) + sizeof(Node))) 等到无穷大.

If you stored a Node object inside itself, what would sizeof(Node) be? It would be sizeof(int) + sizeof(Node), which would be equal to sizeof(int) + (sizeof(int) + sizeof(Node)), which would be equal to sizeof(int) + (sizeof(int) + (sizeof(int) + sizeof(Node))), etc. to infinity.

这样的对象是不可能存在的.这是不可能.

An object like that can't exist. It's impossible.

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