c++ 多重继承强制转换

Question

我有一个类:

class Base;

我还有一个界面

class Interface;

接下来我要创建一个类

class Derived : public Base, public Interface;

如果我有 Base *object = new Derived;

如何将 object 转换为 Interface ？ (当然，如果我知道 object 实际上是派生类)

编辑:

我尝试过 dynamic_cast 和 static_cast(未编译)。 所以让我再解释一下这个问题:

我有:

class Object {...}

class ITouchResponder
{
public:
    virtual bool onTouchBegan(XTouch *touch) = 0;
    virtual void onTouchMoved(XTouch *touch) = 0;
    virtual void onTouchEnded(XTouch *touch) = 0;
};

class Ball : public Object, public ITouchResponder {...};

class TentacleSensor : public Object, public ITouchResponder {...}

对象有一个 bool touchable_ 属性。如果为真，则对象正在实现 ITouchResponder 接口(interface)。

当我使用它时:

bool Level::onTouchBegan(XTouch *touch)
{
    ...
    ITouchResponder *responder = callback.nearestTouchable();
    if (responder)
    {
        if (responder->onTouchBegan(touch))
        {
            if (responder != ball_)
            {
                touch->setUserData(responder);
            }
        }
    }

    return true;
}

ITouchResponder *QueryCallback::nearestTouchable() const
{
    for (list<Object*>::const_iterator it = objects_.begin(); it != objects_.end(); ++it)
    {
        if ( (*it)->isTouchable() ) return (*it)->asTouchResponder();
    }
    return 0;
}

asTouchResponder 是 Object 的一个方法:

    ITouchResponder * Object::asTouchResponder()
    {
        assert(touchable_);
        ITouchResponder *res = dynamic_cast<ITouchResponder*>(this);
        assert(res);
        return res;
    }

我在 xcode 中有严重的过度错误。

但如果我让 Object : public ITouchResponder 一切正常。我做错了什么？

完整的对象类:

class Object// : public ITouchResponder
{
public:
    struct Def
    {
        Def()
        {
            level = 0;
            world = 0;
            touchable = false;
            acceptsContacts = false;
            body = 0;
            node = 0;
        }
        Level *level;
        b2World *world;
        bool touchable;
        bool acceptsContacts;
        b2Body *body;
        XNode *node;
    };

    Object(const Def &def);
    virtual ~Object();

    virtual void update(float dt);
    bool isTouchable() const {return touchable_;}

    void addDependantObject(Object *object);
    void removeDependantObject(Object *object);

    virtual void objectWillBeRemoved(Object *object) {} //this function is automatically called to every dependant object when object is removed

    virtual XVec2 position() const;
    virtual float rotation() const;

    bool acceptsContacts() const {return acceptsContacts_;}

    b2Body *body() const {return body_;}

    Level *level() const {return level_;}
    b2World *world() const {return world_;}

    ITouchResponder *asTouchResponder();
    /*
    virtual bool onTouchBegan(XTouch *touch) {
        return false;
    }
    virtual void onTouchMoved(XTouch *touch)
    {
    }
    virtual void onTouchEnded(XTouch *touch) 
    {
    }*/

protected:
    Level *level_;
    b2World *world_;
    bool touchable_;
    bool acceptsContacts_;

    XNode *node_;
    b2Body *body_;

    list<Object*> dependantObjects_;
};

Answer 1

如果Base有virtual函数(甚至是virtual析构函数)，那么:

Derived *pDerived = dynamic_cast<Derived *>(object);

否则，使用

Derived *pDerived = static_cast<Derived *>(object);

---

请注意，如果 Base 没有虚函数，则 dynamic_cast 将不会编译。在 dynamic_cast 中，只有源必须是多态对象才能编译，如果目标不是多态，则 dynamic_cast 将返回空指针:

假设A和B是多态类型，而C是非多态类型，那么

A *pA = dynamic_cast<A*>(new C()); //error - source is not polymorphic!

A *pA = dynamic_cast<A*>(new B()); //ok
if ( pA == 0 )
      cout << "pA will be null if B is not derived from A" << endl;

C *pC = dynamic_cast<C*>(new B()); //ok
if ( pC == 0 )
       cout << "pC must be null" << endl;