虛函數被類的構造解構函式和成員函數調用虛函數的執行過程

複製代碼 代碼如下:#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;
}

會有這樣的輸出

即使func1是虛函數，在base類和derived的建構函式和解構函式裡面，都是調用自己類裡面的func1。

而在普通成員函數func2調用func1，就會走虛函數的流程。