链表C ++中的赋值运算符 [英] Assignment operator in linked list C++

问题描述

我正在尝试用c ++实现链接列表。

I'm trying to implement linked list in c++.

我这样实现我的赋值运算符：

I implement my assignment operator like this:

// assignment operator
template<class T>
LinkedList<T>& LinkedList<T>::operator = (const LinkedList& rhs) {

if (&rhs != this) {
    Node *tmp = head;

    while (tmp -> next) {
        head = head -> next;
        delete tmp;
        tmp = head;
    }

    tmp = rhs -> head;

    while (tmp) {
        append(tmp);
        tmp = tmp -> next;
    }
}

    return *this;
}

在我的主要功能中，我使用以下代码进行测试：

In my main function, i use the following code to test:

LinkedList<int> *lst1 = new LinkedList<int>(7);

LinkedList<int> *lst2 = new LinkedList<int>(101);

std::cout << lst1 -> head -> data << std::endl;
std::cout << lst2 -> head -> data << std::endl;

lst1 = lst2;

std::cout << lst1 -> head -> data << std::endl;

delete lst1;
delete lst2;            <--------------- error here

期望，控制台输出：

7 101101

7 101 101

但是当程序尝试删除lst2时，我得到一条错误消息：

But when the program try to delete lst2, i get an error saying:

释放的指针未分配

pointer being freed was not allocated

我使用调试器，找出程序何时进行赋值：

I use debugger and find out when the program is doing assignment:

lst1 = lst2;

lst1实际上是指指向lst2的地址，而不是获取lst2的副本，所以当我删除lst1，lst2已经消失。

lst1 is actually referring to the address that points lst2 instead of getting a copy of lst2, so when i delete lst1, lst2 is already gone.

那么有人可以告诉我我的赋值运算符怎么了吗？

很抱歉，如果这是一个新手问题，但我已经花了几个小时而无法弄清。

I'm sorry if this is a novice question but I've been spending a few hours and could not figure out.

我完成的代码如下所示：

My completed code is shown below:

template<class T>
class LinkedList {

private:
    class Node {
    public:
        T data;
        Node *next;

        // default constructor
        Node() = default;

        // constructor with data
        Node(const T& data) : data(data), next(NULL) {}

    };

public:
    Node *head;

    LinkedList(const LinkedList& copyLst);
    LinkedList& operator=(const LinkedList& byValList);
    LinkedList() : head(NULL){}
    LinkedList(Node *newNode) : head(newNode) {}
    LinkedList(T val) {
        head = new Node(val);
    }
    ~LinkedList();

    static LinkedList<int> sumLists(const LinkedList<int>& lst1, const LinkedList<int>& lst2) ;

    void insertAtFront(T val);
    void insertAtEnd(T val);
    void printList();
    void insert(T val);

    void append(const Node&);
};

// copy constructor
template<class T>
LinkedList<T>::LinkedList(const LinkedList<T>& copyLst) {

    const Node *cpCurrent = copyLst.head;
    Node *lsCurrent = NULL;

    if (cpCurrent != NULL) {
        head = new Node(cpCurrent -> data);
        lsCurrent = head;

        cpCurrent = cpCurrent -> next;

    }

    while (cpCurrent != NULL) {
        Node *newNode = new Node(cpCurrent -> data);
        lsCurrent -> next = newNode;

        lsCurrent = lsCurrent -> next;
        cpCurrent = cpCurrent -> next;
    }
}

// assignment operator
template<class T>
LinkedList<T>& LinkedList<T>::operator = (const LinkedList& rhs) {

    if (&rhs != this) {
        Node *tmp = head;

        while (tmp -> next) {
            head = head -> next;
            delete tmp;
            tmp = head;
        }

        tmp = rhs -> head;

        while (tmp) {
            append(tmp);
            tmp = tmp -> next;
        }
    }

    return *this;
}

// destructor
template<class T>
LinkedList<T>::~LinkedList() {
    Node *current = head;

    while (current != NULL) {
        head = head -> next;
        delete current;
        current = head;
    }
}

template<typename T>
void LinkedList<T>::append(const Node& node ){

    if (NULL == head) {
        Node *newNode = new Node(node -> data);
        head = newNode;
    } else {
        Node *current = head;

        while (current -> next) {
            current = current -> next;
        }


        Node *newNode = new Node(node -> data);
        current -> next = newNode;
    }
}

推荐答案

您的当前实现会复制复制构造函数中已经存在的代码，所以为什么不重用它呢？

Your current implementation duplicates code that already exists in the copy constructor, so why not reuse it?

如果您有一个有效的复制构造函数和析构函数，请使用复制/交换惯用法是实现赋值运算符的最简单，最安全的方法。

If you have a working copy constructor and destructor, usage of the copy / swap idiom would be the easiest and safest way to implement the assignment operator.

#include <algorithm>
//...
template<class T>
LinkedList<T>& LinkedList<T>::operator = (const LinkedList<T>& rhs) 
{
    LinkedList<T> temp(rhs);
    std::swap(temp.head, head);
    return *this;
}

鉴于您的复制构造函数和析构函数可以正常工作，因此可以保证正常工作。我们创建对象 rhs 的副本 temp 并将其内容替换为的内容* this 。当 temp 在返回时被销毁时，它会同时包含以前在 * this 中的旧数据。

Given that your copy constructor and destructor work correctly, this is guaranteed to work correctly. We create a copy temp of the object rhs and swap out its contents with the contents of *this. When temp gets destroyed at the return, it takes along with it the old data that used to be in *this.

如果您使用的是C ++ 11编译器，则可以利用带有传递值的传入参数进行移动构造。

If you have a C++ 11 compiler, you can take advantage of move construction on the passed-in parameter with pass-by-value.

#include <algorithm>
//...
template<class T>
LinkedList<T>& LinkedList<T>::operator = (LinkedList<T> rhs) 
{
    std::swap(rhs.head, head);
    return *this;
}

*请注意，您需要交换的所有成员使用复制和交换习惯用法的班级，否则您可能会使班级成员的不变量*

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