虚函数被类的构造析构函数和成员函数调用虚函数的执行过程

复制代码 代码如下:

#include<iostream>

 

class base{
public:

    base()
    {
        std::cout<<std::endl;
        std::cout<<"base constructor"<<std::endl;
        func1();
        std::cout<<std::endl;
    }

    virtual ~base()
    {
        std::cout<<std::endl;
        std::cout<<"base distructor"<<std::endl;
        func1();
        std::cout<<std::endl;
    }
    virtual void func1()
    {
        std::cout<<"base virtural func1"<<std::endl;
    }

    void func2()
    {
        std::cout<<"base member func2"<<std::endl;
        func1();
        std::cout<<std::endl;
    }
};

class derived:public base{
public:
    derived()
    {
        std::cout<<std::endl;
        std::cout<<"derived constructor"<<std::endl;
        func1();
        std::cout<<std::endl;
    }

    virtual ~derived()
    {
        std::cout<<std::endl;
        std::cout<<"derived distructor"<<std::endl;
        func1();
        std::cout<<std::endl;
    }

    virtual void func1()
    {
        std::cout<<"derived virtual func1"<<std::endl;
    }

};

int main()
{
    base *point = new derived();
    point->func2();
    delete point;
    return 0;
}

 

会有这样的输出

即使func1是虚函数，在base类和derived的构造函数和析构函数里面，都是调用自己类里面的func1。

而在普通成员函数func2调用func1，就会走虚函数的流程。