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(shù),在base類和derived的構(gòu)造函數(shù)和析構(gòu)函數(shù)里面,都是調(diào)用自己類里面的func1。
而在普通成員函數(shù)func2調(diào)用func1,就會走虛函數(shù)的流程。
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