调用对象方法传递超类期望子类

Question

例如我有三个班级

class Shape
{
  static bool collide(Box * a, Box * b)
  static bool collide(Box * a, Sphere * b)
  static bool collide(Sphere * a, Sphere * b)
};

class Box : public Shape{};

class Sphere : public Shape{};

例如在其他地方我有这段代码

Shape * a = new Box();
Shape * b = new Sphere();

Shape::collide(a, b);

编译器引发错误。

如何在不添加太多逻辑的情况下自动向下转换作为参数给出的对象？

Answer 1

编译器不能自动转换对象，你必须手动转换它们，例如：

class Box;
class Sphere;

class Shape
{
  virtual ~Shape(){}

  static bool collide(Shape *a, Shape *b);
  static bool collide(Box *a, Box *b);
  static bool collide(Box *a, Sphere *b);
  static bool collide(Sphere *a, Sphere *b);
};

class Box : public Shape{};

class Sphere : public Shape{};

bool Shape::collide(Shape *a, Shape *b)
{
  if (Box *a_box = dynamic_cast<Box*>(a))
  {
    if (Box *b_box = dynamic_cast<Box*>(b))
      return collide(a_box, b_box);

    if (Sphere *b_sphere = dynamic_cast<Sphere*>(b))
      return collide(a_box, b_sphere);
  }
  else if (Sphere *a_sphere = dynamic_cast<Sphere*>(a))
  {
    if (Sphere *b_sphere = dynamic_cast<Sphere*>(b))
      return collide(a_sphere, b_sphere);
  }
  return false;
}

bool Shape::collide(Box *a, Box *b) { ... }
bool Shape::collide(Box *a, Sphere *b) { ... }
bool Shape::collide(Sphere *a, Sphere *b) { ... }

Shape * a = new Box();
Shape * b = new Sphere();

Shape::collide(a, b);

不用说，随着新形状的添加，这会变得有点乏味。您应该改为将单个 virtual 方法添加到 Shape 本身，并让派生类覆盖它以根据需要处理不同的类型，例如：

class Shape
{
  virtual ~Shape(){}
  virtual bool collide(Shape * b) { return false; };
};

class Box : public Shape{
  bool collide(Shape * b) override {
    if (Box *b_box = dynamic_cast<Box*>(b)) {
      return ...;
    }
    if (Sphere *b_sphere = dynamic_cast<Sphere*>(b)) {
      return ...;
    }
    return Shape::collide(b);
  };
};

class Sphere : public Shape{
  bool collide(Shape * b) override { 
    if (Sphere *b_sphere = dynamic_cast<Sphere*>(b)) {
      return ...;
    }
    return Shape::collide(b);
  };
};

Shape * a = new Box();
Shape * b = new Sphere();

a->collide(b);