先看错误代码(无virtual版本)

#include <iostream>
using namespace std;

class base1{
public:
     void show()const{
        cout<<"showing base1"<<endl;
    }
};

class base2: public base1{
public:
     void show()const{
        cout<<"showing base2"<<endl;
    }
};

class derived: public base2{
public:
     void show()const{
        cout<<"showing derived"<<endl;
    }
};
void show(base1*i){
    i->show();
}
int main(){
    base1 b1;
    base2 b2;
    derived d;
    show(&b1);
    show(&b2);
    show(&d);





    return 0;
}

这是因为没有virtual
结果是

加了virtual关键字之后

#include <iostream>
using namespace std;

class base1{
public:
    virtual void show()const{
        cout<<"showing base1"<<endl;
    }
};

class base2: public base1{
public:
     void show()const{
        cout<<"showing base2"<<endl;
    }
};

class derived: public base2{
public:
     void show()const{
        cout<<"showing derived"<<endl;
    }
};
void show(base1*i){
    i->show();
}
int main(){
    base1 b1;
    base2 b2;
    derived d;
    show(&b1);
    show(&b2);
    show(&d);





    return 0;
}

结果是
原因：show函数需要在编译时候就确定，但是因为我们调用的时候不能确定，所以要在base1那里加个virtual（动态绑定）

ps:写void show(base1*i)的时候竟然写成了base1&i,深感code力下降qaq