带有纯虚函数的C++模板返回值

Question

我有一个抽象的 Handle 类，它包含对 T 类型对象的引用。我希望能够将该类转换为 Handle，其中 U 是 T 的父类(super class)。我会使用继承，但这在这里不起作用。我该怎么做呢？什么是好的替代品？

示例伪代码:

template<class T>
class Handle {
public:
    virtual ~Handle () {}
    virtual T & operator* () const = 0;
    virtual T * operator-> () const = 0;
    virtual template<class U> operator Handle<U>* () const = 0; // being lazy with dumb pointer
};

template<class T>
class ConcreteHandle : public Handle<T> {
public:
    explicit template<class U> ConcreteHandle (U * obj) : obj(obj) {}
    virtual ~ConcreteHandle () {}
    virtual T & operator* () const {
        return *obj;
    }
    virtual T * operator-> () const {
        return obj;
    }
    virtual template<class U> operator Handle<U>* () {
        return new ConcreteHandle<U>(obj);
    }
private:
    T * obj;
};

---

按照要求，这就是我正在做的

class GcPool {
public:
    virtual void gc () = 0;
    virtual Handle<GcObject> * construct (GcClass clazz) = 0;
};

class CompactingPool : public GcPool {
public:
    virtual void gc () { ... }
    virtual Handle<GcObject> * construct (GcClass clazz) { ... }
private:
    Handle<GcList<Handle<GcObject> > > rootSet; // this will grow in the CompactingPool's own pool
    Handle<GcList<Handle<GcObject> > > knownHandles; // this will grow in the CompactingPool's own pool.
};

knownHandles 需要与 Handle 兼容，以便它可以在 CompatingPool 的 rootSet 中。 rootSet 也是如此。我将引导这些特殊句柄，这样就不会出现先有鸡还是先有蛋的问题。

Answer 1

virtual template<class U> operator Handle<U>* () const  =0;

语言规范不允许使用模板虚函数。

考虑 this code at ideone ，然后看到编译错误:

error: templates may not be ‘virtual’

---

现在你能做什么？一种解决方案是:

template<class T>
class Handle {
public:

    typedef typename T::super super; //U = super, which is a superclass of T.

    virtual ~Handle () {}
    virtual T & operator* () const = 0;
    virtual T * operator-> () const = 0;

    //not a template now, but still virtual
    virtual super operator Handle<super> () const = 0;  
};

即在派生类中定义一个基类的typedef，并在Handle中使用。像这样:

struct Base {//...};

struct Derived : Base { typedef Base super; //...};

Handle<Derived>  handle; 

---

或者你可以定义特征，如:

struct Base {//... };

struct Derived : Base { //... };

template<typename T> struct super_traits;

struct super_traits<Derived>
{
   typedef Base super;
};

template<class T>
class Handle {
public:

    typedef typename super_traits<T>::super super; //note this now!

    virtual ~Handle () {}
    virtual T & operator* () const = 0;
    virtual T * operator-> () const = 0;

    //not a template now, but still virtual
    virtual super operator Handle<super> () const = 0; 
};

在我看来，super_traits 是一个更好的解决方案，因为您可以在不编辑派生类的情况下定义它们的特征。此外，您可以根据需要定义任意数量的 typedef；假设你的派生类有多个基类，你可能想定义很多 typedef，或者最好是 typelist .