C ++向下转换到基于变量的派生类

问题描述

假设我有一个基类Shape，派生类Triangle，Square和Circle。
Shape的成员是一个intshapeType。

Suppose I have a base class "Shape", and derived classes "Triangle", "Square", and "Circle".A member of "Shape" is an int "shapeType".

如果shapeType == 1，那么它是一个三角形。
如果shapeType == 2，那么它是一个正方形。
如果shapeType == 3，那么它是一个圆。

If shapeType==1, then it's a triangle.If shapeType==2, then it's a square.If shapeType==3, then it's a circle.

我有兴趣知道，我只有一个Shape对象，对象，如果有一种方法通过使用shapeType值动态下降到适当的派生类。

I'm interested in knowing given I have just a "Shape" object that was once a derived-object, if there is a way to "dynamically" down-cast to the proper derived class by using the shapeType value.

我知道我可以做一个硬代码switch语句，大致如下：

I know I can do a hard-code switch statement, roughly like:

Triangle* t;
Square* s;
Circle* c;

switch (shape->shapeType) {
case 1:
   t = (Triangle*)shape;
case 2: 
   ...
}

上面需要我做一个指针的EVERY派生类的可能性。我想知道是否有一种方法来做这个没有硬编码每个类，而是以某种方式确定类类型映射，其中关键是shapeType和值是类类型。

However, the above requires me to make a pointer of EVERY derived class possibility. I am wondering if there is a way to do this without hard-coding every class, but instead somehow determine the class type map where the key is the shapeType and the value is the class type.

推荐答案

如果他们有虚函数，则使用 dynamic_cast ：

If they've virtual functions, then use dynamic_cast:

t = dynamic_cast<Triangle*>(shape);
if ( t )
{
     //use t
}

但是请注意：你应该尝试以这样一种方式定义类和虚函数，你几乎不需要使用 dynamic_cast 。一般来说，喜欢定义良好的接口和多态性。

But take a note: you should try defining the classes and virtual functions in such a way that you would hardly need to use dynamic_cast. Prefer well-defined interface, and polymorphism, in general.

以下是一个示例，

class Shape
{
   public:
     ~virtual Shape() {} //destructor must be virtual - important!
      virtual double Area() const = 0;
};

class Triangle : public Shape
{
   public:
     Triangle(double a, double b, double c);
     virtual double Area() const 
     {
         //calculate area and return it!
     }
};

Shape *s = new Traingle(10,20,30);
double aread = s->Area(); //calls Traingle::Area()

不需要使用 shapeType 变量。

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