Question

我不知道为什么PrintList（）不会终止。它是一个LinkedList，所以当我转到下一个时，它应该终止。当我执行AddNode一次然后打印时，终止，当我执行addNode两次时，print不会终止。

我在构造函数中创建5个空白点的原因是因为我需要在程序启动时创建这5个空白点。

此外，如果我添加两次，如何指定指向第二个值的指针？

#pragma once
class LinkedList
{
private:
    typedef struct node {
        int data;
        node* next;
    }* nodePtr;

    nodePtr n;

    nodePtr head;
    nodePtr curr;
    nodePtr temp;
public:
    LinkedList();
    void AddNode(int addData);
    void PrintList();
    ~LinkedList();
};

#include "LinkedList.h"
#include<cstdlib>
#include<iostream>
using namespace std;


LinkedList::LinkedList()
{
    head = NULL;
    curr = NULL;
    temp = NULL;

    n = new node;

    for (int x = 1; x <= 5; x++) {
        //cout << n<<endl;
        n->next = n;
    }

}

void LinkedList::AddNode(int addData) {
    //nodePtr n = new node;
    n->next = NULL;
    n->data = addData;
    cout << n <<endl;
    if (head != NULL) {
        curr = head;
        while (curr->next != NULL) {
            curr = curr->next;
        }
        curr->next = n;
    }
    else {
        head = n;
    }
}


void LinkedList::PrintList() {
    curr = head;
    while (curr != NULL) {
        cout << curr->data << endl;
        curr = curr->next;
    }

}


LinkedList::~LinkedList()
{
    head = NULL;
    curr = NULL;
    temp = NULL;
    delete n;
}

#include <iostream>
#include <fstream>
#include <string>
#include <cstdlib>
#include "LinkedList.h"
using namespace std;

int main() {
    LinkedList *l = new LinkedList();
    l->AddNode(5);
    l->AddNode(8);
    l->PrintList();
    system("pause");
    return 0;
}

Answer 1

由于您未将n设置为其他节点，因此节点n始终相同因此，当您执行n-＆gt; next和n-＆gt;数据时，每次都会修改相同的节点

void LinkedList::AddNode(int addData) {
    //nodePtr n = new node; // you need to uncomment this
    n->next = NULL;
    n->data = addData;
    cout << n <<endl;
    if (head != NULL) {
        curr = head;
        while (curr->next != NULL) {
            curr = curr->next;
        }
        curr->next = n;
    }
    else {
        head = n;
    }
}


After your first addNode(5), lets examine the values.
head = n
head->data = 5
head->next = null

After your second addNode(8)
head = n
head->data = 8
head->next = n // set by "curr->next = n" .

所以你这里有问题。当您尝试循环链接列表时，它将变为head-＆gt; next-＆gt; head-＆gt; next-＆gt; head-＆gt; next-＆gt; head-＆gt; next .....导致无限环

打印功能不会终止

1 个答案: