Question

我创建了一个链接列表，当我添加到列表的前面或后面时，它可以正常工作。我打印出来，一切都很好看。然后我从列表的前面（pop_front）中删除，由于某种原因，行temp = head->getPrevious();（List.cpp）返回值0x000000，所以很明显它实际上并没有获得指向前一个节点的指针。我已经介入它，我看不到任何东西。任何帮助都会很棒！

Main.cpp的

#include "stdafx.h"
#include "List.h"



int main()
{

    List partsList;

    partsList.push_front(22);
    partsList.push_front(25);
    partsList.push_front(32);
    partsList.push_back(100);

    partsList.display();
    cout << "now we are going to remove the first item in the list" << endl;
    system("PAUSE");

    partsList.pop_front();

    partsList.display();

    system("PAUSE");
    cout << "now we are going to remove the LAST item from the list" << endl;

    partsList.pop_back();

    partsList.display();

    system("PAUSE");
    return 0;
}

List.h

#pragma once
#include "node.h"
#include <iostream>
using namespace std;
class List
{

private:
    int listSize;
    Node* n;
    Node* temp;
    Node* head;
    Node* tail;

public:
    List();
    void push_front(int);
    void push_back(int);
    void pop_front();
    void pop_back();
    void display();

    ~List();
};

List.cpp

#include "stdafx.h"
#include "List.h"


List::List()
{
}

void List::push_front(int dat)
{
    if (listSize == 0) {

        n = new Node;
        n->setData(dat);
        listSize++;
        temp = n;
        head = n;
        tail = n;

    }
    else {

        n = new Node;
        n->setData(dat);
        listSize++;
        temp = head;
        head = n;
        n->setNext(temp);
        n->setPrevious(nullptr);
        temp->setPrevious(n);
        temp = n;
    }

}

void List::push_back(int dat)
{
    if (listSize == 0) {

        n = new Node;
        n->setData(dat);
        listSize++;
        temp = n;
        head = n;
        tail = n;
    }
    else {

        n = new Node;
        n->setData(dat);
        listSize++;

        temp = tail;
        temp->setNext(n);
        n->setPrevious(temp);
// SET NEXT TO NULL
        temp = n;
        tail = temp;
    }
}

void List::pop_front()
{
    temp = head->getPrevious();
    delete head;
    head = temp;
    listSize--;

}

void List::pop_back()
{
    temp = tail->getPrevious();
    delete tail;
    tail = temp;
    tail->setNext(nullptr);
    listSize--;

}

void List::display()
{

    Node* test = head;
    for (int i = 0; i < listSize; i++) {
        cout << test->getData() << endl;
        test = test->getNext();
    }

}


List::~List()
{
}

Node.h

#pragma once
class Node
{
private:
    int data;
    Node* next;
    Node* previous;

public:
    Node();
    int getData();
    void setData(int);
    void setNext(Node*);
    void setPrevious(Node*);
    Node* getPrevious();
    Node* getNext();
    ~Node();
};

Node.cpp

#include "stdafx.h"
#include "Node.h"


Node::Node()
{
}

int Node::getData()
{
    return data;
}

void Node::setData(int dat)
{
    data = dat;
}

void Node::setNext(Node* nextNode)
{
    next = nextNode;
}

void Node::setPrevious(Node* prev)
{
    previous = prev;
}

Node * Node::getPrevious()
{
    return previous;
}

Node * Node::getNext()
{
    return next;
}





Node::~Node()
{
}

Answer 1

在head->getPrevious();中，您正在调用nullptr，显然正在返回getNext()，并将其设置为列表的新头。它应该是{{1}}。

顺便说一句：你不应该在List对象中存储n和temp等临时变量。在操作方法之外不需要它们，只会使对象膨胀。

链接列表，没有返回上一个

1 个答案: