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Virtual destruct（虚析构函数）

面试的时候，我答了虚函数是怎么工作的，但是当面试官问我为什么析构函数的基类一定需要virtual时候，我答的不是太好，所以来总结下
先说结论，当基类的指针指向子类的对象时，如果delete此指针，如果基类没有virtual析构函数，那么调用时只会调用基类的析构函数，而子类的空间未释放。
如以下例子

class BaseClass
{
private:
    int *m_Basepointer;
public:
    BaseClass() { std::cout << "it's Base Construct!" << std::endl; }
    ~BaseClass()
    {
        delete m_Basepointer;
        std::cout << "it's Base Deconstruct!" << std::endl;
    }
};

class SubClass:public BaseClass
{
public:
    SubClass() { std::cout << "it's Sub Construct!" << std::endl; }
    ~SubClass() 
    {
        std::cout << "it's Sub Deconstruct!" << std::endl;
    }

};

int main()
{
    BaseClass *p_b = new BaseClass;
    delete p_b;
    std::cout << "---------------------------------\n";
    SubClass *p_s = new SubClass;
    delete p_s;
    std::cout << "---------------------------------\n";
    BaseClass *p_b_s = new SubClass;
    delete p_b_s;

    std::cin.get();
}

从如上例子可以看出子类的析构函数未被调用，此时造成内存泄露。但是在基类加上virtual后，如下

#include<iostream>

class BaseClass
{
private:
    int *m_Basepointer;
public:
    BaseClass() { std::cout << "it's Base Construct!" << std::endl; }
    virtual ~BaseClass()
    {
        delete m_Basepointer;
        std::cout << "it's Base Deconstruct!" << std::endl;
    }
};

class SubClass:public BaseClass
{
public:
    SubClass() { std::cout << "it's Sub Construct!" << std::endl; }
    ~SubClass() 
    {
        std::cout << "it's Sub Deconstruct!" << std::endl;
    }

};

int main()
{
    BaseClass *p_b = new BaseClass;
    delete p_b;
    std::cout << "---------------------------------\n";
    SubClass *p_s = new SubClass;
    delete p_s;
    std::cout << "---------------------------------\n";
    BaseClass *p_b_s = new SubClass;
    delete p_b_s;

    std::cin.get();
}

成功释放，这就是为何要在基类加上virtual防止内存泄露，保证安全释放。

Virtual destruct（虚析构函数）

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原文地址：https://www.cnblogs.com/Yekko/p/12516539.html