协变虚拟函数和智能指针 [英] Covariant virtual functions and smart pointers

问题描述

在C ++中，子类可以在重写虚拟函数时指定不同的返回类型，只要该返回类型是原始返回类型的子类(并且两者均作为指针/引用返回)即可.

In C++, a subclass can specify a different return type when overriding a virtual function, as long as the return type is a subclass of the original return type (And both are returned as pointers/references).

是否可以将此功能扩展到智能指针?(假设智能指针是某些模板类)

Is it possible to expand this feature to smart pointers as well? (Assuming a smart pointer is some template class)

说明:

class retBase {...};
class retSub : public retBase {...};

class Base
{
    virtual retBase *f();
};

class Sub : public Base
{
    virtual retSub *f();     // This is ok.
};


class smartBase
{
    virtual smartPtr<retBase> f();
};

class smartSub : public smartBase
{
    virtual smartPtr<retSub> f();     // Can this be somehow acheived?
};

编辑:正如Konrad Rudolph所建议的那样，这不可能直接实现.但是，我遇到了这种方法:

As Konrad Rudolph suggested, this is not directly possible. However, I ran accross this method:

class smartBase
{
    protected:
        virtual retBase *f_impl();
    public:
        smartPtr<refBase> f()
        {
             return f_impl();
        }
};

class smartSub : public smartBase
{
    protected:
        virtual retSub *f_impl();
    public:
        smartPtr<refSub> f()
        {
             return f_impl();
        }
};

您建议这样做吗?

推荐答案

是否可以将此功能扩展到智能指针?(假设智能指针是某些模板类)

Is it possible to expand this feature to smart pointers as well? (Assuming a smart pointer is some template class)

否:C ++不知道/不允许协变或逆变模板.即使 A 继承自 B ，类型 Ptr< A> 和 Ptr< B> 之间也没有关系.

No: C++ doesn't know/allow covariant or contravariant templates. There's no relation between types Ptr<A> and Ptr<B>, even if A inherits from B.

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