有没有办法upcast到一个抽象类，而不是修改它每次一个类派生自它？ [英] Is there no way to upcast into an abstract class and not modify it each time a class is derived from it?

问题描述

#include<iostream>
using namespace std;

class Abs
{
        public:
        virtual void hi()=0;
};

class B:public Abs
{
        public:
        void hi() {cout<<"B Hi"<<endl;}
        void bye() {cout<<"B Bye"<<endl;}
};
class C:public Abs
{
        public:
        void hi() {cout<<"C Hi"<<endl;}
        void sayonara() {cout<<"C Sayonara"<<endl;}
};

int main()
{
        Abs *bb=new B;
        bb->bye();
        Abs *cc=new C;
        cc->sayonara();
}//main

编译器说

test2.cpp: In function ‘int main()’:
test2.cpp:26: error: ‘class Abs’ has no member named ‘bye’
test2.cpp:28: error: ‘class Abs’ has no member named ‘sayonara’

因为这个问题，我每次创建一个新的派生类继承时，我必须添加功能Abs类（向上转换是强制的我要做的我计划要求它是如此）。我不想触摸基类创建后。
不是这个问题违反了原则，一旦你做一个基类，你不必修改它。任何方式来解决这个问题？
ps：我已经看到了工厂设计模式和原型设计模式，但是他们似乎都无法解决它。

Because of this problem, I'll have to add functions to the Abs class each time I create a new derived class which inherits from it (Upcasting is compulsory for me to do. The program I'm planning requires it to be so). I don't want to touch the base class once it's created. Doesn't this problem violate the principle that once you make a base class, you won't have to modify it ever. Any way to resolve this problem? p.s: I've seen the factory design pattern and the prototype design patterns, but both of them can't seem to be able to solve it.

推荐答案

好吧，我不知道你想要什么（以及为什么你想要这样），但：

Well, i'm not sure to understand exactly what you want (and why you want it that way) but:

int main()
{
        Abs *bb=new B;
        static_cast<B*>(bb)->bye();
        Abs *cc=new C;
        static_cast<C*>(cc)->sayonara();
}//main

将工作。

你只需要确保 bb 在你之前真的是 B * c> static_cast 。

You just have to be sure that bb is really a B* before you static_cast.

您也可以使用 dynamic_cast 如果 bb 的类型不正确。

You may also use dynamic_cast which will return a null pointer if bb is not of the correct type.

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