如果没有列表本身就可以执行链接列表 [英] It is possible and how to do a linked list without the list itself

问题描述

大家好，

所以我的问题是这样的：

有可能以及如何在没有列表的情况下进行链接列表？



我的意思是我在网上找到的每个链表（我可能错过了很多）都是由两部分组成的，Node类（或struct）和一个列表类包含链接节点的列表，管理所有prev和next对象。所以我的问题是我是否只能从Node类中获得链表。所以Node类应该有* prev和* next，并且应该管理* prev和* next对象的分配和解除定位。这甚至是可管理的并且是个好主意吗？

Hi everyone,
So my question is this:
It is possible and how to do a linked list without the list itself?

What I mean by this is on every linked list I can find on the net (I may have missed many) is constructed from two parts, the Node class (or struct) and a list class which containts the list of the linked nodes, managing all of the prev and next objects. So my question is can I have a "linked list" only from the Node class. So the Node class should have *prev and *next and should manage the allocation and dellocation of the *prev and *next objects. Is that even managable and a good idea?

推荐答案

是的，在纯粹的C ++中非常简单 - 你只需使用指针。

如何使用C / C ++创建链接列表 [ ^ ]

Yes, in "pure" C++ that's pretty simple - you just use pointers.
How to create Linked list using C/C++[^]

虽然我感谢你以不同的方式思考，但我会说没有'list'的链接'list'并不是一个好主意。链表基本上是链接在一起的节点列表。如果没有列表类，您打算在哪里保留对链接列表的头部和/或结尾的引用？您计划在哪里实现链接列表特定的功能，如遍历，插入，删除等？最终没有'List'，你最终会用C ++实现一个面向过程的链接列表程序。



'OriginalGriff'共享的链接显示了a的实现纯C中的链表。虽然理解链表的基础是一个很好的开始，但它不是用C ++实现链表的最佳方式。

While I appreciate you thinking in a different way, I would say that is not a good idea to have a linked 'list' without 'list'. A linked list is basically a list of Nodes linked together. When there is no 'List' class where do you plan to keep the reference to the Head and/or End of the linked list? And where do you plan to implement the Linked List specific functionalities such as traversal, insertion, deletion etc.? Without a 'List' eventually you would end up in implementing a procedural oriented linked list program in C++.

The link that 'OriginalGriff' has shared shows the implementation of a linked list in "pure C". While it is a good start to understand the basics of linked list, it is not the best way to implement a linked list in C++.

引用：

Node类应该有* prev和* next，并且应该管理* prev和* next对象的分配和解除分配。

the Node class should have *prev and *next and should manage the allocation and de-allocation of the *prev and *next objects.

当然，prev和next指针应该是Node类的一部分。但它不是Node类为它们分配内存的责任。实际上* prev和* next分别指向将节点链接到其上一个节点和下一个节点的指针。如果Node本身负责分配和取消分配其prev和next指针，那么它就像一个Node创建其上一个和下一个节点。在这种情况下，最基本的问题是谁负责创建第一个节点？

Of course the prev and next pointers should be a part of the Node class. But its not the duty of the Node class to allocate memory for them. Actually *prev and *next are pointers to link a Node to its previous and next Nodes respectively. If a Node itself is responsible for allocating and de-allocating its prev and next pointers, then it would be like a Node creating its previous and next Nodes. In that case the most basic question would be 'Who is responsible for creating the first Node?'

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