Question

我在c ++中使用模板来创建链表。我编写了一些函数，但删除节点（从头部/尾部）无法正常工作。它显示错误并终止。在调试时，这个＆＃39;在监视窗口中显示包含一个null的头，我不明白为什么。这是我的代码：

SLlistSc.h

#ifndef SLlistSc_H
#define SLlistSc_H

#include<iostream>
using namespace std;

template<class T> class SLlist;

template <class T> class snode{
    friend class SLlist<T>;
private:
    T info;
    snode<T> *next;
public:
    inline T getter() const { return info; }
    inline void setter(T setValue){ info = setValue; }
    snode(){ info = 0; next = 0; }
    snode(T inf , snode *nex = 0){ info = inf; next = nex; }
    ~snode(){ delete next; }

};
template <class T> class SLlist{
private:
    snode<T> *head, *tail;
    static int total;
public:
    SLlist(){ head = tail = 0; }

    ~SLlist(){
        snode<T> *p = head;
        while (head != 0){
            head = head->next;
            delete p;
            p = head;
        }
    }
    inline void incrcnt(){ total++; }
    inline void decrcnt(){ total--; }
    void displayList();
    void addToHead(T inf);
    void addToTail(T inf);
    T deleteFromHead();
    T deleteFromTail();
    int returnIndex();
    void deleteInfo();
    void deleteAll();
};

template <class T> void SLlist<T>::displayList(){
    snode<T> *p = head;
    while (p != 0){
        cout << "\n->" << p->info;
        p = p->next;
    }
    return;
}

template <class T> void SLlist<T>::addToHead(T inf){
    snode<T> *temp = new snode<T>(inf, 0);
    if (head == 0){ head = tail = temp; temp = 0; delete temp; incrcnt(); return; }
    else { temp->next = head; head = temp; temp = 0; delete temp; incrcnt(); return; }
}

template <class T> void SLlist<T>::addToTail(T inf){
    snode<T> *temp = new snode<T>(inf, 0);
    if (head == 0){ head = tail = temp; temp = 0; delete temp; incrcnt(); return; }
    else{ tail->next = temp; tail = temp; temp = 0; delete temp; return; }
}

template <class T> T SLlist<T>::deleteFromHead(){
    if (head == 0){ cout << "\nList is already empty"; return (T)0; }
    if (head == tail){ delete head; head = tail = 0; T info = head->info; return info; }
    else {
        T info = head->info;
        snode<T> *temp = head;
        head = head->next;
        temp = 0;
        delete temp;
        return info;
    }
}

template <class T> T SLlist<T>::deleteFromTail(){
    if (head == 0){ cout << "\nList is already empty"; return (T)0; }
    if (head == tail){ delete head; head = tail = 0; T info = head->info; return info; }
    else {
        T info = tail->info;
        snode<T> *temp = head;
        while (temp != 0)
        {
            temp = tail;
        }
        delete tail;
        tail = temp;
        temp = 0; delete temp;
        return (T)info;
    }
}

#endif

和cpp文件：SLlistSc.cpp

// SLlistSc.cpp : Defines the entry point for the console application.
//

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

template <class T> int SLlist<T>::total = 0;

void main()
{
    cout << "\t\t\t Singly Linked List Super Class Implementation";
    SLlist<int> List1;
    int userchoice=0, inf=0;
    do{
        cout << "\n\n\t\t Menu"
            << "\n\t1. Display List"
            << "\n\t2. Add to Head"
            << "\n\t3. Add to tail"
            << "\n\t4. Delete from Head"
            << "\n\t5. Delete from Tail"
            << "\n\t6. Exit";
        cin >> userchoice;

        switch (userchoice){
        case 1: List1.displayList(); break;
        case 2: cout << "\nEnter info to be added:";
                cin >> inf;
                List1.addToHead(inf); break;
        case 3: cout << "\nEnter info to be added:";
                cin >> inf;
                List1.addToTail(inf); break;
        case 4: inf = List1.deleteFromHead();
                cout << "\n Value" << inf << "deleted from list"; break;
        case 5: inf = List1.deleteFromTail();
                cout << "\n Value" << inf << "deleted from list"; break;
        case 6: cout << "\n\t\t\t\tExiting";
                exit(0);
        default: continue;
        }
    } while (userchoice < 4);
}

我想问的另一件事是，当我们在一个类（例如A类）上使用模板时，我们需要在任何地方用A替换A.是因为编译器生成了一个新类，即。

A＆LT; T＆GT; isnt = A.

在snode类中使用友元类SLlist时，为什么我必须声明如下：

template<class T> class SLlist;

在snode课程之前？为什么我不能直接在snode类中声明它：

friend class SLlist<T>;

如果我的代码中有可能改进，请继续努力。感谢。

Answer 1

您正在尝试使用无效的指针，因为已释放包含该指针的块。

Microsoft C ++调试运行时库使用模式0xFEEEFEEE覆盖释放的内存，这使得此问题更容易识别。

删除链表中的节点不起作用？

1 个答案: