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。