Class template specialization in class scope?

Question

Why is the specialization S in A legal and S in B not?

( if B is not commented out ) GCC 4.8.1: error: explicit specialization in non-namespace scope ‘class B’

#include <type_traits>
#include <iostream>

class Y {};
class X {};

struct A {
  template<class T, class = void>
  class S;

  template<class T>
  struct S < T, typename std::enable_if< std::is_same< Y, T >::value >::type > 
  {
    int i = 0;
  };

  template<class T>
  struct S < T, typename std::enable_if< std::is_same< X, T >::value >::type > 
  {
    int i = 1;
  };
};

/*
class B
{
    template<class T>
    class S;

    template<>
    class S < Y > {};

    template<>
    class S < X > {};
};
*/


int main()
{
    A::S< X > asd;
    std::cout << asd.i << std::endl;
}

on coliru: B commented out

on coliru: with B (error)

Quimper

You can have partial specializations in non-namespace scope, but not explicit specializations. – Loom

@Loom Could you elaborate? This answer does not seem to agree. – Verla

Answer 1

The comments by @jrok pretty much explains your compiler error. Nested classes in general, and nested class templates in particular, are a dusty corner of the language that you can easily avoid (taking to heart Sutter's advice "Write what you know and know what you write").

Simply make a namespace detail to define your class templates SA and SB and their specializations and then define a nested template type alias S inside both A and B

namespace detail {

  template<class T, class = void>
  class SA;

  template<class T>
  struct SA < T, typename std::enable_if< std::is_same< Y, T >::value >::type > 
  {
    int i = 0;
  };

  template<class T>
  struct SA < T, typename std::enable_if< std::is_same< X, T >::value >::type > 
  {
    int i = 1;
  };

  template<class T>
  class SB;

  template<>
  class SB < Y > {};

  template<>
  class SB < X > {};
}

struct A
{
    template<class T>
    using S = detail::SA<T>;
};

struct B
{
    template<class T>
    using S = detail::SB<T>;
};

Granted, for this case it might seem overkill, but if you ever want to make A and B class templates themselves, and specialize A and B, then you can only specialize the nested class templates if you also specialize the enclosing class. In short: just avoid these issues altogether by an extra level of compile-time indirection.

Aplite

I believe that the using statements are incorrect, and should look like this: template<class T> using S = detail::SA<T>; See: en.cppreference.com/w/cpp/language/type_alias – Insinuate