简单的 C++ 链表

Simple C++ Linked List(简单的 C++ 链表)

问题描述

我以前有很多在 Java 中使用链表的经验，但我似乎对 C++ 中的这种简单尝试感到困惑.我在运行时遇到分段错误，据我所知，这与分配空指针有关，但我对解决方案不知所措.

感谢大家非常有帮助的回复.代码现在可以工作了，但是尝试在 linksList::addNode 的末尾使用

delete p;

会导致运行时出现分段错误.只是好奇是否有人知道这是为什么?

这是我更新的代码:

#include 使用命名空间标准；类节点{上市:整数数据；节点 * 下一个；节点(int x){数据 = x;下一个 = NULL;}节点(int x，节点* y){数据 = x;下一个 = y;}};类链表{节点*头；上市:链接列表(){头 = NULL;}无效添加节点(整数值){节点 *p;如果(头 == NULL)head = 新节点(值，NULL)；别的{p=头；while(p->next !=NULL)p=p->下一个；p->next = 新节点(值，NULL)；}}无效打印(){节点 * p;p = 头；而(p ！= NULL){cout<<p-＞数据＜＜"
";p=p->下一个；}}};int main(void){链表测试；test.addNode(4);test.addNode(76);test.addNode(12);测试打印()；返回(0)；}

首先，在 linkedList::addNode 方法中，你有构造 if (head = NULL), 这将结束 assigning 给 head;您需要 == 运算符.

二、关于线路:

head = &(Node (value, NULL));

由于一些不直观的原因，这行不通.您将获得对 Node 的引用，但该节点将在方法结束后立即超出范围，并且尝试引用它会导致分段错误.您需要使用 new 运算符(与其他类似行相同):

head = new Node(value, NULL);

如果您添加删除节点的方法，请确保delete该节点，然后它不会像在 Java 中那样自动进行垃圾收集.

<块引用><块引用>

侧边栏: 想想会发生这样的事情:当您执行 Node(value, NULL) 时，您正在使用一个声明如下的临时变量:

Node hiddenTempNode(value, NULL);

这不会在堆栈上以外的任何地方为对象分配空间——这与在堆栈上为 int 和 Node * 分配空间非常相似.变量.结果，只要你离开方法，对象就消失了，指向它的指针在使用时会做一些奇怪的事情.

第三，注意:您可能希望在单参数构造函数中设置 next = NULL，以确保它始终具有值.与您的默认构造函数类似.

第四:你的 linkedList::print 方法循环直到 p->next 是 NULL 并打印 的值p->下一个;如果您想获取第一个和最后一个项目，p->next 的那些出现可能应该更改为 p.

I have plenty of previous experience with linked lists in Java, but I seem to have confused myself with this simple attempt in C++. I am getting a segmentation fault at runtime, which from what I understand has to do with assigning a null pointer, but I am at a loss for a solution.

Edit: Thank you all for the very helpful responses. The code is now working, but trying to use

delete p;

at the end of linkedList::addNode results in a segmentation fault at runtime. Just curious if anyone knew why that is?

Here is my updated code:

#include <iostream>
using namespace std;

class Node{
    public:
        int data;
    Node * next;
    Node(int x){
        data = x;
        next = NULL;
        }
    Node(int x, Node * y){
        data = x; 
        next = y;
        }
    };


class linkedList{
Node *head;
public:
    linkedList(){
        head = NULL;
        }
    void addNode(int value){
        Node *p;
        if(head == NULL)
            head = new Node (value, NULL);
        else{
            p=head;
            while(p->next !=NULL)
                p=p->next;
            p->next = new Node (value, NULL);
            }
        }
    void print(){
        Node * p;
        p = head;
        while(p != NULL){
            cout << p->data << "
";
            p = p->next;
            }
        }
};


int main(void){
linkedList test;
test.addNode(4);
test.addNode(76);
test.addNode(12);
test.print();
return(0);
}

First, in linkedList::addNode method, you have the construction if (head = NULL), which will wind up assigning to head; you want the == operator.

Second, about the line:

head = &(Node (value, NULL));

For somewhat unintuitive reasons, this won't work. You'll get a reference to a Node, but that node will go out of scope as soon as the method ends, and attempts to reference it will lead to a segmentation fault. You need to use the new operator (same with the other similar line):

head = new Node(value, NULL);

If you add a method for removing a node, make sure to delete the node then—it won't get automatically garbage-collected like it will in Java.

Sidebar: Think of what happens like this: when you do Node(value, NULL), you're using a temporary variable that's declared like this:

Node hiddenTempNode(value, NULL);

This doesn't allocate space for an object anywhere except on the stack—it's very similar to allocating space for an int and a Node * on the stack as separate variables. As a result, as soon as you leave the method, the object disappears and the pointer to it will do weird things when used.

Third, beware: you may want to set next = NULL in your single-parameter constructor, to ensure that it always has a value. Similarly for your default constructor.

Fourth: your linkedList::print method is looping until p->next is NULL and printing the value of p->next; those occurrences of p->next should probably be changed to just p if you want to get the first and last items.

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