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Question

I've got a class template with a template template parameter, and I want to declare this parameter (that is, all of its specializations) as a friend . But I can't find the correct syntax.

template <template <class> class T>
struct Foo {

    template <class U>
    friend T;           // "C++ requires a type specifier for all declarations"

    template <class U>
    friend struct T;    // "declaration of 'T' shadows template parameter"

    template <class U>
    friend struct T<U>; // "cannot specialize a template template parameter"

    pretty<please>
    lets(be) friends T; // Compiler shook its standard output in pity
};

How can I declare a template template parameter as friend ?

Coliru snippet

Answer 1

I've found this problem is very interesting to research. According to standard (C++11 and above) this should works fine, I guess:

template <template <class> class U>
struct identity {
    template <typename T>
    using type = U<T>;
};

template <template <class> class U>
struct Foo {
    template <typename>
    friend class identity<U>::type;
};

Due to one indirection level, there are no any shadowing and name reusing here, and this situation described as following:

cppreference :

Both function template and class template declarations may appear with the friend specifier in any non-local class or class template (...). In this case, every specialization of the template becomes a friend, whether it is implicitly instantiated, partially specialized, or explicitly specialized. Example: class A { template<typename> friend class B; } class A { template<typename> friend class B; }

However, it seems like clang and MSVC think otherwise: see coliru or godbolt . Moreover, their error messages seem meaningless, so it looks like a compiler bug (or just an incorrectly detected syntax error). GCC compiles and executes mentioned snippets perfectly, but I still not sure it's a correct solution.

Answer 2

Templates tend to get programmers to over-complicate things. I'm not exactly sure what you are trying to do, and I suspect that there is probably a better way to do it, but this should work.

template <template <class> class T, class U>
class Foo {
    friend class T<U>;
};