通过基类指针将派生类引用传递给函数 [英] Passing a Derived class reference to a function through a Base Class Pointer

问题描述

我很确定这是OOP 101(也许是102?)，但是我在理解如何执行此操作时遇到了一些麻烦.

I'm pretty sure this is OOP 101 (maybe 102?) but I'm having some trouble understanding how to go about this.

我试图在项目中使用一个函数，根据传递给它的对象来产生不同的结果.从我今天所读的书中，我相信我已经找到了答案，但是我希望这里的人可以肯定我.

I'm trying to use one function in my project to produce different results based on what objet is passed to it. From what I've read today I believe I have the answer but I'm hoping someone here could sure me up a little.

//Base Class "A"
class A
{
    virtual void DoThis() = 0; //derived classes have their own version
};

//Derived Class "B"
class B : public A
{
    void DoThis() //Meant to perform differently based on which
                  //derived class it comes from
};

void DoStuff(A *ref) //in game function that calls the DoThis function of
{ref->DoThis();}     //which even object is passed to it.
                     //Should be a reference to the base class

int main()
{
    B b;

    DoStuff(&b);     //passing a reference to a derived class to call
                     //b's DoThis function
}

有了这个，如果我有多个从Base派生的类，我是否可以将任何Derived类传递给DoStuff(A *ref)函数，并利用来自Base的虚函数?

With this, if I have multiple classes derived from the Base will I be able to pass any Derived class to the DoStuff(A *ref) function and utilize the virtuals from the base?

我正确执行此操作吗?还是我离开这里了?

Am I doing this correctly or am I way off base here?

推荐答案

因此，利用Maxim共享给我的IDEOne(非常感谢)，我能够确认自己做得正确

So, utilizing IDEOne which was shared with me by Maxim (Thank you very much), I was able to confirm that I was doing this correctly

#include <iostream>
using namespace std;

class Character 
{
public:
    virtual void DrawCard() = 0;    
};

class Player: public Character
{
public:
    void DrawCard(){cout<<"Hello"<<endl;}
};

class Enemy: public Character
{
public:
    void DrawCard(){cout<<"World"<<endl;}
};

void Print(Character *ref){
    ref->DrawCard();
}

int main() {

    Player player;
    Enemy enemy;

    Print(&player);

    return 0;
}

正如我希望的那样，

Print(&player)和Print(&enemy)确实调用了它们各自的DrawCard()函数.这无疑为我打开了一些大门.谢谢那些帮助过的人.

Print(&player) and Print(&enemy) do call their respective DrawCard() functions as I hoped they would. This has definitely opened some doors to me. Thank you to those who helped.

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