Question

下面是没有汇编的,我如何能够这样做? 我认为,这个例子表明了我的意图,但如果有人混淆,我就试图增加一个模糊之处。

template<typename T>
class A
{
private:
    struct B
    {
        template<typename T2> 
        B& operator=( const A<T2>::B& right ){} //  how can I make this work?
    };

    template<typename T2> friend class A;
    template<typename T2> friend class A<T2>::B;
};

Answer 1

您的派任经营人的基本想法存在缺陷,因为即使添加了<条码>的类型<> /代码>,该代码为不可替代背景<>。因此,模板参数将永远不会被推卸,除非你明确指明如下类型:<条码>B.operator=<有些——类型与“其他_B”。

An easier version would be to just make it a normal function template, and SFINAE your way out.

#include <type_traits> template<class> struct is_a_B; template<class B2> typename std::enable_if< is_a_B<B2>::value, B& >::type operator=(B2 const& other){ // ... }

现在所有左边都是<代码>is_a_B。海峡。你们能够以可能的虚假积极态度,使这种情况很容易发生:

template<class B> struct is_a_B{ typedef char yes; typedef yes (&no)[2]; template<class T> static yes test(typename T::I_am_a_B_type*); template<class T> static no test(...); static bool const value = sizeof(test<B>(0)) == sizeof(yes); };

Just provide the I_am_a_B_type 页: 1

http://ideone.com/MYMiH”rel=“nofollow> 关于Ideone的活例子。评论b1 = 5;行,并汇编成。

And now for the ~~slightly more~~ ~~perverted~~ complicated way with no false-positives. :)

template<bool Cond, class OnTrue> struct and_v{ static bool const value = OnTrue::value; }; template<class OnTrue> struct and_v<false, OnTrue>{ static bool const value = false; }; template<class B> struct is_a_B{ typedef char yes; typedef yes (&no)[2]; template<class T> static yes has_parent(typename T::parent*); template<class T> static no has_parent(...); template<class T> static yes is_A(A<T>*); static no is_A(...); template<class T> struct lazy_test{ typedef typename std::add_pointer<typename T::parent>::type p_type; static bool const value = sizeof(is_A(p_type(0))) == sizeof(yes); }; static bool const value = and_v<sizeof(has_parent<B>(0)) == sizeof(yes), lazy_test<B>>::value; };

For this one you need a typedef A<T> parent; inside B. It s staged in two parts:

First I test if a parent typedef exists, and if it does

If it s actually a typedef of the A class template.

可悲的是,逻辑操作者(&&,>,?: t t 如同我所希望的那样,在模板代码上不出现短路,因此,我不得不写上<代码>和_v的模板+ a lazy tester,只要有即 类型def>,只有经过评估。

http://ideone.com/tyVTo”rel=“nofollow> 关于Ideone的活例子。 http://ideone.com/szZsc” rel=“nofollow”

Answer 2

It should not work at all. Here is the reason

template<typename T>
class A
{
    private:
    struct B
    {
        template<typename T2> 
        B& operator=( const typename A<T2>::B& right ){}
    };

    template<typename T2> friend class A;
    template<typename T2> friend class A<T2>::B;
};

template <>
class A<int>
{
};

各位都可以看到: A< int >:B 班次!

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