虛函數被類的構造析構函數和成員函數調用虛函數的執行過程

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的構造函數和析構函數裡面，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。