Why is partial specialziation of a nested class template allowed, while complete isn't?

Question

    template<int x> struct A {                                                                                                    
        template<int y> struct B {};.                                                                                             
        template<int y, int unused> struct C {};                                                                                  
    };                                                                                                                            

    template<int x> template<> 
    struct A<x>::B<x> {}; // error: enclosing class templates are not explicitly specialized

    template<int x> template<int unused> 
    struct A<x>::C<x, unused> {}; // ok

So why is the explicit specialization of a inner, nested class (or function) not allowed, if the outer class isn't specialized too? Strange enough, I can work around this behaviour if I only partially specialize the inner class with simply adding a dummy template parameter. Makes things uglier and more complex, but it works.

I would consider complete specializations as a subset of partial specializations - especially because you can express every complete specialization as a partial with adding a dummy parameter. So this disambiguation between partial and full specialization doesn't really make sense for me.

Unfortunatly nobody at comp.std.c++ dared to answer, so I am putting it up here again with a bounty.

Note: I need this feature for recursive templates of the inner class for a set of the outer class and the specialization of the inner parameter does depend on the outer template parameter.

Answer 1

C++ Standard explicitly prohibits full specialization of member template classes in first case. According to 14.7.3/18:

In an explicit specialization declaration for a member of a class template or a member template that appears in namespace scope, the member template and some of its enclosing class templates may remain unspecialized, except that the declaration shall not explicitly specialize a class member template if its enclosing class templates are not explicitly specialized as well.

Answer 2

My guess as to why this happens: complete specializations are no longer "template classes/functions", they are are "real" classes/methods, and get to have real (linker-visible) symbols. But for a completely-specialized template inside a partially-specialized one, this would not be true. Probably this decision was taken just to simplify the life of compiler-writers (and make life harder for coders, in the process :P ).

Answer 3

Backing up Virgil's argument (he was faster than I posting the same rationale), consider this:

template<typename T1>
class TOuter
  {
  public:
    template<typename T2>
    class TInner
      {
      public:
        T1 m_1;
        T2 m_2;
      };
  };

template<typename T1>
template<>
class TOuter<T1>::TInner<float>
  {
  public:
    T1    m_1;
    float m_2;
 };

Is TInner fully specialized, or partially specialized because of T1?

Edit:

After considering some of the other comments - it seems that you want to have a full specialization based on the template parameter in the outer class. If you nest the inner class's implementation, that seems to work in Visual Studio 2005:

template<typename T1>
class TOuter
  {
  public:
    template<typename T2>
    class TInner
      {
      public:
        std::string DoSomething() { return "Inner - general"; }
        T2 m_2;
      };

    template<>
    class TInner<T1>
      {
      public:
        std::string DoSomething() { return "Inner - special"; }
        T1 m_1;
      };
  };

TOuter::TInner will correctly be the specialization of TInner. I could not get it to compile with the implementation outside of the template.

Answer 4

You can work around this behavior by delegating the real work to another structure though:

namespace detail
{
  template <class T, class U>
  struct InnerImpl {};
}

template <class T>
struct Outer
{
  template <class U>
  struct Inner: detail::InnerImpl<T,U>
  {
  };
};

Now you can specialize InnerImpl as you wish