继承多态函数调用

Question

class Shape 
{
   public:
      virtual void draw() const {cout<<"draw shape"<<endl;}
};



class Point : public Shape 
{
   public:
      Point( int a= 0, int b = 0 ) {x=a; y=b;}  // default constructor
      int getx() const {return x;}
      int gety() const {return y;}
      virtual void draw() const {cout<<"draw point("<<x<<","<<y<<")\n";}
   private:
      int x, y;   // x and y coordinates of Point
};


class Circle : public Point 
{
   public:  // default constructor
      Circle( double r = 0.0, int x = 0, int y = 0 ):Point(x,y) {radius=r;}
      virtual void draw() const 
      {cout<<"draw circle("<<getx()<<","<<gety()<<","<<radius<<")\n";}
   private:
      double radius;   // radius of Circle
};

void functionCall(Shape *arrayOfShapes[3])
{
    Shape shape;
    Point point( 7, 11 );            // create a Point
    Circle circle( 3.5, 22, 8 );     // create a Circle

    arrayOfShapes[0] = &shape;
    arrayOfShapes[1] = &point;
    arrayOfShapes[2] = &circle;

}

int main()
{


    Shape *arrayOfShapes[3];

    functionCall(arrayOfShapes);
    for(int i=0; i<3; ++i)
    arrayOfShapes[i]->draw();

    return 0;
}

当我尝试运行时，出现了分段错误。似乎主函数无法检索arrayOfShapes[3] 对象？

有没有办法调用传入对象指针的函数并在完成后返回对象指针？

Answer 1

您不能像这样在局部函数中创建形状，因为将局部变量的地址放入数组会使它们在其范围之外可用：

Shape shape;
arrayOfShapes[0] = &shape; // <<== Pointer to local

但是，您可以这样做：

arrayOfShapes[0] = new Shape;
arrayOfShapes[1] = new Point( 7, 11 );
arrayOfShapes[2] = new Circle( 3.5, 22, 8 );

这种方法在动态内存中创建形状，允许在函数返回时使用它们。

注意：尽管Circle 需要一个原点，但圆绝对不是一个点。因此，这个声明在逻辑上是不合理的：

class Circle : public Point // This implies that Circle is a Point

虽然您可以争辩说一个点是一个半径为零的圆，但像这样构造继承也是一个坏主意。更好的方法是使Circle 包含 Point 作为其来源：

class Circle : public Shape 
{
   public:
      const Point origin;
      // default constructor
      Circle( double r = 0.0, int x = 0, int y = 0 ):origin(x,y),
 radius(r) {}
      virtual void draw() const 
      {cout<<"draw circle("<<origin.getx()<<","<<origin.gety()<<","<<radius<<")\n";}
   private:
      double radius;   // radius of Circle
};