C++模板化返回值与纯虚函数

Question

我有一个抽象的Handle <T>类,它包含类型为T的对象的引用.我希望能够将该类转换为Handle <U>,其中U是T的超类.我会使用继承,但这不起作用.我该怎么做呢？什么是好的选择？

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;
};

---

根据要求,这就是我正在做的事情

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,因此它可以在CompatingPool的rootSet中.rootSet也是如此.我将引导这些特殊的手柄,这样就不会出现鸡和蛋的问题.

Answer 1

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

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语言规范不允许模板虚函数。

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考虑ideone 中的这段代码，然后查看编译错误：

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错误：模板可能不是 \xe2\x80\x98virtual\xe2\x80\x99

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现在你能做什么？一种解决方案是这样的：

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template<class T>\nclass Handle {\npublic:\n\n    typedef typename T::super super; //U = super, which is a superclass of T.\n\n    virtual ~Handle () {}\n    virtual T & operator* () const = 0;\n    virtual T * operator-> () const = 0;\n\n    //not a template now, but still virtual\n    virtual super operator Handle<super> () const = 0;  \n};\n

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即typedef在派生类中定义基类的a，并在Handle. 像这样的东西：

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struct Base {//...};\n\nstruct Derived : Base { typedef Base super; //...};\n\nHandle<Derived>  handle; \n

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或者您可以定义特征，如下所示：

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struct Base {//... };\n\nstruct Derived : Base { //... };\n\ntemplate<typename T> struct super_traits;\n\nstruct super_traits<Derived>\n{\n   typedef Base super;\n};\n\ntemplate<class T>\nclass Handle {\npublic:\n\n    typedef typename super_traits<T>::super super; //note this now!\n\n    virtual ~Handle () {}\n    virtual T & operator* () const = 0;\n    virtual T * operator-> () const = 0;\n\n    //not a template now, but still virtual\n    virtual super operator Handle<super> () const = 0; \n};\n

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在我看来，super_traits这是一个更好的解决方案，因为您可以定义派生类的特征而不对其进行编辑。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，你可能想要定义许多 typedef，或者最好是一个typelist。

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