为什么地址不一样？

本文介绍了为什么地址不一样？的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我已经声明了一个基类和一个派生类，它将基类作为虚拟公共继承。之后我宣布一个派生类的对象，然后声明一个指向已声明类的对象的基类的指针对象

I have declared a base class and a derived class which inherit base class as virtual public . After that I am declaring a oject of derived class and then declaring a pointer object of base class pointing to object of declared class

class A
{
    protected:
    int m_A;

public:
    A(int a=0):m_A(a){ cout << "A "<< m_A<<endl;}
    ~A(){cout << "~A"<<endl;}
    void Display(){ cout << "I am in A :D"<<endl;}


};

class D1: virtual public A
{

protected:
    int m_D1;

public:

    D1(int b=0, int a=0):m_D1(b),A(a)
    {cout << "D1 "<<m_D1<<endl;}
    ~D1(){cout << "~D1"<<endl;}
    void Display(){cout <<" I am in D1 :D "<<endl;}

};
int main()
{



   D1 d(1,2);
   d.Display();
   A *a=&d;
   cout << "&d= "<<&d<<"  a= "<<a<<endl;
   a->Display();

}

我的问题是为什么& d和a的值（地址）在这种情况下不一样，但是如果我们不继承基类是虚拟的那么它们是相同的吗？

My question is why the values(address) of "&d" and "a" are not same in this case , however they are same if we do not inherit base class is virtual ?

and

为什么地址不一样？

问题描述

推荐答案