c++ - 删除链表中的节点不起作用？

Question

我正在使用 C++ 中的模板来创建链表。我已经编写了一些函数，但是删除节点(从头/尾)无法正常工作。它显示错误并终止。调试时，监 window 口中显示的“this”包含一个空头，我不知道为什么。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，当我们在类(例如类 A)上使用模板时，我们需要在所有地方将 A 替换为 A。是因为编译器生成了一个新类即。

A<T> isnt = A.

而在 snode 类中使用 friend 类 SLlist 时，为什么我必须声明如下:

template<class T> class SLlist;

在 snode 类之前？为什么我不能直接在 snode 类中将其声明为:

friend class SLlist<T>;

此外，如果我的代码有改进的机会，请加倍努力。谢谢。

Answer 1

您正试图使用​​一个无效的指针，因为包含该指针的 block 已被释放。

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。