C++ 多重继承，虚析构确定

来个例子就明了了：

#include<iostream>
using namespace std;
class Base1
{
public:
    virtual ~Base1(){
        cout<<"Base1"<<endl;
    }
};
class Base2
{
public:
    virtual ~Base2(){
        cout<<"Base2"<<endl;
    }
};
class D1:public Base1{
public:
    ~D1()
    {
        cout<<"D1"<<endl;
    }
};
class D2:public Base2{
public:
    ~D2()
    {
        cout<<"D2"<<endl;
    }
};
class MI:public D1,public D2{
public:
    ~MI()
    {
        cout<<"MI"<<endl;
    }
};
int main(int argc, char *argv[])
{
    Base1 *pb1=new MI;
    Base2 *pb2=new MI;
    D1 *pd1=new MI;
    D2 *pd2=new MI;
    delete pb2;
    cout<<endl;
    delete pd1;
    cout<<endl;
    delete pd2;
    cout<<endl;
    return 0;
}

特别关注delete pd2；

实质上它的结果和pd1的释放是一模一样的，虽然pd2是D2类型指针，但是析构会整个析构MI对象，这个对象包含了Base1子对象，D1子对象，Base2子对象，D2子对象，还有自身的部分。所有的过程为逆序析构，打印结果为：