使用不在基类中的派生方法 [英] Using derived methods that aren't in the base class

问题描述

我正在做一个Shape类，它允许你获得某个形状的positionX，positionY和区域。派生类是一个Circle，它有一个附加的半径​​变量。

I'm doing a Shape class that allows you to get the positionX, positionY, and area of some shape. The derived class is a Circle, which has an added variable of radius.

class Shape
{
public:
    Shape( double positionX, double positionY );
    virtual double getPositionX() = 0;
    virtual double getPositionY() = 0;
    virtual double area() = 0;

 class Circle : public Shape
{
public:
    Circle( double positionX, double positionY, double radius );
    virtual double getPositionX();
    virtual double getPositionY();
    virtual double area();
    virtual double getRadius();

在我的主要内容中，我有一个Shapes数组，第一个对象是一个新的Circle并且正在调用这个圈子上的方法。

In my main, I have an array of Shapes with the first object being a new Circle and am calling methods on this Circle.

int main()
{
    Shape* s[2];

    s[0] = new Circle( 2.0, 3.0, 4.0 );
    cout << s[0]->getPositionX() << endl; // Works fine.
    cout << s[0]->getPositionY() << endl; // Works fine.
    cout << s[0]->getRadius() << endl; // Doesn't work because s is a shape and my shape class has                             
                                       // no getRadius() method.

如何调用getRadius（）方法？

How do I call the getRadius() method?

推荐答案

Ug ......一个好的设计完全避免了这一点。如果你需要得到某个东西的半径，那么最好是能够提供半径的东西。如果形状是三角形，你的程序应该做什么？

Ug...A good design avoids this entirely. If you need to get the radius of something, then it better be something that can provide a radius. If the shape was a triangle, what should your program do?

现在解决这个问题的方法是使用动态强制转换：

Now the way around this is to use dynamic cast:

    Circle* circ = dynamic_cast<Circle*>(s[0]);
    if (circ)//check if cast succeeded
    {
       cout << circ->getRadius() << endl;
    }

但这段代码闻起来。一般来说，如果你必须使用dynamic_cast（这很慢而且很混乱）那么你应该重新考虑你的设计。

But this code smells. In general, if you have to use dynamic_cast (which is slow and messy) then you should be rethinking your design.

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