C++新手：析构函数

发布于 2024-12-23 05:28:06 字数 1021 浏览 0 评论 0原文

我只是创建一个简单的列表，然后销毁它。出了问题，我总是收到这个恼人的错误消息：

Expression: _BLOCK_TYPE_IS_VALID(pHead->nBlockUse)

这是代码：

#include<iostream>
#include<Windows.h>
using namespace std;

struct node
{
    int data;
    node *next;
};

class list
{
protected:
    node *top;
public:
    list()
    {
        top=NULL;
    }

    list random()
    {
        int x=rand()%10;
        for(int i=0; i<x; i++)
        {
            node *p=new node;
            p->data=rand()%100;
            p->next=top;
            top=p;
        }
        return *this;
    }

    void show()
    {
        for(node *p=top; p; p=p->next)
        {
            cout<<p->data<<" ";
        }
        cout<<"\n";
    }

    ~list()
    {
        node *r;
        for(node *p=top; p; p=r)
        {
            r=p->next;
            delete p;
        }
    }
};

int main()
{
    srand(GetTickCount());
    list a;
    a.random().show();
    return 0;
}

原文

I'm just creating a simple list and then destroying it. And something is going wrong and I always get this annoying error message:

Expression: _BLOCK_TYPE_IS_VALID(pHead->nBlockUse)

Here's the code:

#include<iostream>
#include<Windows.h>
using namespace std;

struct node
{
    int data;
    node *next;
};

class list
{
protected:
    node *top;
public:
    list()
    {
        top=NULL;
    }

    list random()
    {
        int x=rand()%10;
        for(int i=0; i<x; i++)
        {
            node *p=new node;
            p->data=rand()%100;
            p->next=top;
            top=p;
        }
        return *this;
    }

    void show()
    {
        for(node *p=top; p; p=p->next)
        {
            cout<<p->data<<" ";
        }
        cout<<"\n";
    }

    ~list()
    {
        node *r;
        for(node *p=top; p; p=r)
        {
            r=p->next;
            delete p;
        }
    }
};

int main()
{
    srand(GetTickCount());
    list a;
    a.random().show();
    return 0;
}

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冷…雨湿花 2024-12-30 05:28:06

这：

list random()

应该是：

list &random()

原因是您的版本返回实例 a 的副本，并且该副本在调用 show() 后被破坏..并且破坏会破坏与 a 使用的内存相同。如果您确实想让random() 返回一个副本，则需要实现一个复制构造函数，该构造函数对a 具有的内部列表进行深层复制。

This:

list random()

should be:

list &random()

The reason is that your version returns a copy of your instance a, and that copy get destructed after show() is called.. and that destruction destroys the same memory as that a is using. If you really want to have random() return a copy, you need to implement a copy constructor that does a deep copy of the internal list that a has.

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給妳壹絲溫柔 2024-12-30 05:28:06

您的问题是您正在复制 list 但没有定义复制构造函数。隐式定义的复制构造函数只会复制 top 指针，因此您最终会尝试删除同一节点链两次。

当您从按值返回 *this 副本的 random() 成员函数返回 *this; 时，就会发生复制。

最短的修复方法是通过在类的私有部分声明复制构造函数和复制赋值运算符来使您的类不可复制。

private:
    list(const list&);
    list& operator=(const list&);

然后你可以让random返回void，似乎没有一个很好的理由为什么它也制作一个副本。

然后，您可以这样调用它：

    list a;
    a.random();
    a.show();

较长的解决方法是通过完整实现 list(const list&) 和 list& 来使您的 list 可复制; operator=(const list&) 正确复制正在复制的源 list 的所有节点。

Your problem is that you are copying your list but you don't define a copy constructor. The implicitly defined copy constructor will just copy the top pointer so you end up attempting to delete the same chain of nodes twice.

The copy occurs when you return *this; from your random() member function returning a copy of *this by value.

The shortest fix would be to make your class non-copyable by declaring a copy constructor and copy assignment operator in the private section of your class.

private:
    list(const list&);
    list& operator=(const list&);

You can then make random return void, there doesn't seem to be a good reason why it makes a copy as well.

You could then just call it like this:

    list a;
    a.random();
    a.show();

The longer fix would be to make your list copyable by making a full implementation of list(const list&) and list& operator=(const list&) that correctly duplicates all the nodes of the source list being copied.

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