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。