c++中为什么析构函数要被设置为虚函数(virtual)

1.如果类A可能会被继承，那么A的析构函数一定要被设置为virtual虚函数；

　原因：如果基类的析构函数不是虚函数，当创建一个基类指针指向一个派生类对象，当释放此基类指针时，子类的析构函数不会被调用；

　如下：

class CFather
{
public:
    CFather();
    virtual ~CFather();
    
public:
    virtual void test();
    //virtual void getHouse() = 0;
    
public:
    int b;
    static const int c = 5;
};

CFather::CFather()
{
    infof("CFather...create! g_test:%d
", g_test++);
}

CFather::~CFather()
{
    infof("~CFather...destory
");
}

void CFather::test()
{
    infof("CFather...test
");
}

class CChild : public CFather
{

public:
    CChild();
    ~CChild();
public:

    void test();
    virtual void getHouse();
private:
    int e;
};

CChild::CChild()
{
    infof("CChild...create:g_test:%d
", g_test++);
}

CChild::~CChild()
{
    infof("~CChild...destory
");
}

void CChild::test()
{
    infof("CChild...test
");
}

void CChild::getHouse()
{
    infof("CChild getHouse
");
}

int main()
{
    Infra::setPrintLogLevel(6);

    CFather * cpt = new CChild;

    delete cpt;
        
    return 0;
}

输出结果：

CFather...create! g_test:100

CChild...create:g_test:20



 ~CChild...destory

~CFather...destory