协助单链表(c++)程序

Assistance with Singly Linked List (C++) program

本文关键字:c++ 程序 链表 单链表      更新时间:2023-10-16

我已经开始用c++学习链表,我已经写了一个程序来创建和显示列表。它编译得很好,但是执行失败。

代码如下:

#include <iostream>
using namespace std;
class node
{
    public:
    int data;                       // information contained by the node
    node *next;                     // a pointer pointing to an instance of the class node.
};
class list
{
public:
    node *head;
    node *temp, *tp;
    void createlist();
    void displaylist();
};
void list :: createlist()
{
    int size;
    cout<<"Enter the size of the list you want to create.n";
    cin>>size;
    temp=head;
    for(int k=1;k<=size;k++)
    {
    if(head == NULL)
    {
        head = new node[1];
        cout<<"Enter the data for node number 1n";
        cin>>head->data;
        head->next=NULL;
    }
    else
    {
            temp->next=new node[1];
            temp=temp->next;
            cout<<"Enter the data for node number n"<<k;
            cin>>temp->data;
            temp->next=NULL;
    }
    }
}

void list::displaylist()
{
    if (head==NULL)
    {
        cout<<"List is empty.";
    }
    else
    {
        while(tp!=NULL)
        {
        tp = head;
        cout<<tp->data;
        tp = tp->next;
        }
    }
}
int main()
{
        list obj;
    cout<<"Creating list...n";
    obj.createlist();
    cout<<"Displaying list...n";
    obj.displaylist();
    return 0;
}

#include <iostream>
using namespace std;
class node
{
    public:
    int data;                       // information contained by the node
    node *next;                     // a pointer pointing to an instance of the class node.
};
class list
{
public:
    node *head;
    node *temp, *tp;
    void createlist();
    void displaylist();
};
void list::createlist()
{
    int size;
    cout<<"Enter the size of the list you want to create.n";
    cin>>size;
    temp=head;
    for(int k=1;k<=size;k++)
    {
    if(head == NULL)
    {
        head = new node[1];
        cout<<"Enter the data for node number 1n";
        cin>>head->data;
        head->next=NULL;
    }
    else
    {
            temp->next=new node[1];
            temp=temp->next;
            cout<<"Enter the data for node number n"<<k;
            cin>>temp->data;
            temp->next=NULL;
    }
    }
}

void list::displaylist()
{
    if (head==NULL)
    {
        cout<<"List is empty.";
    }
    else
    {
        while(tp!=NULL)
        {
        tp = head;
        cout<<tp->data;
        tp = tp->next;
        }
    }
}
int main()
{
        list obj;
    cout<<"Creating list...n";
    obj.createlist();
    cout<<"Displaying list...n";
    obj.displaylist();
    return 0;
}

输出如下:
Creating list... Enter the size of the list you want to create. 3 Enter the data for node number 1 1

因此,第二个节点的数据不被接受。我是链表的新手,想尝试用这种方法解决问题,我在这上面花了太多时间。我哪里做错了?

数据成员head和tp在对象obj创建时未初始化,并且具有任意值。所以程序有未定义的行为。此外,将临时变量作为类的tp或临时数据成员也是一个坏主意。它们必须是使用它们的成员函数的局部变量。

没有任何意义,因为你正在做分配一个节点数组。

head = new node[1];
/...
temp->next=new node[1];

写简单的

head = new node();
/...
temp->next=new node();

最好将类node定义为类list的内部类。

并且不要忘记写析构函数来释放分配的内存。