5

I have a class template

template< typename G, int N > class Foo { /* ... */ };

I want the specialization for N=0 to be a friend of another class, but I don't know the syntax for it (and I could not find it out myself). I tried:

template< typename T >
class Bar {
  template< typename G > friend class Foo< G, 0 >;

  /* ... */
};

I want for any type G Foo< G, 0 > to be a friend of class Bar< T >. What is the correct syntax for this?

Thank you!

Sh4pe
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2 Answers2

2

In C++03 that is not possible; the C++ standard 14.5.3/9 says the following:

Friend declarations shall not declare partial specializations.

As noted in another answer, there may be some workarounds for this issue, but the specific functionality you are asking for is not available in that standard.

Luckily, C++11 is quite well supported nowadays, and with the ability to specify template aliases, we can achieve just this:

template <typename, typename> struct X{};

template <typename T> 
struct Y
{
    template <typename U> using X_partial = X<T, U>;
    template <typename> friend class X_partial;
};
nikolas
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    This compiles with gcc, but fails to compile with clang version 3.8. The error message is `error: redefinition of 'X_partial' as different kind of symbol template friend class X_partial;` . – ScootyPuff May 04 '14 at 22:22
  • That error you are seeing on clang should be related to not being able to use typedefs in friend statement: https://stackoverflow.com/questions/392120/why-cant-i-declare-a-friend-through-a-typedef – gg99 Jun 28 '18 at 10:51
1

Without C++11 I think the best you can do is a fake type alias, that may require some code (constructor) duplicatation (which may not solve the real problem you're attempting):

template< typename G, int N > class Foo { /* ... */ };

template<typename G> class FooAlias : public Foo<G, 0> { };

template< typename T >
class Bar {
  template< typename G > friend class FooAlias;

  /* ... */
};
Mark B
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