specialize a member template without specializing its parent

Question

I have a class template nested inside another template. Partially specializing it is easy: I just declare another template< … > block inside its parent.

Answer 1

I tend not to use nested classes too much. My main complaint is that they have a tendency to bloat the code of the class they are nested in.

I would therefore propose another workaround:

namespace detail
{
  template <class X, class Z> class BImpl;
  template <class X, class Z> class BImpl<X, A<Z> > {};
  template <class X> class BImpl<X,int> {};
}

template <class X>
class A
{
  template <class Z> struct B: BImpl<X,Z> {};
};

Just note that it requires to pass X as an argument to BImpl if ever you wish to also specialize A. Funny thing is that in this case, I end up with only partial specialization!

Answer 2

Atleast this works in VC 2010. But, I am unable to write the def. of fun() for "int" outside the class declaration. EDIT: Unfortunately g++ has also compilations issues. EDIT: The code below worked on VC 2010.

template<typename X>
class A
{
public:
    A()
    {

    }

    template<typename Y>
    struct B
    {
        void fun();
    };

    template<>
    struct B<int>
    {
        void fun()
        {
            cout << "Specialized version called\n";
        }
        //void fun();
    };



public:

    B<X> b;
};



template<typename X>
template<typename Y>
void A<X>::B<Y>::fun()
{
    cout << "templated version called\n";
}

int main()
{
   A<int> a;
    a.b.fun();
    A<float> a1;
    a1.b.fun();
}

Answer 3

It is illegal under C++ standard 14.7.3/18:

.... the declaration shall not explicitly specialize a class member template if its enclosing class templates are not explicitly specialized as well.

Answer 4

Complex stuff. Your initial code ICE's VC10 Beta2, nice.

First off, I think you have this backwards:

template<> 
template< class X > 
struct A< X >::B< int > {};

X is a template param to struct A, and B is the one fully specialized, so I think it should be this:

template< class X > 
template<> 
struct A< X >::B< int > {};

But even this fails to compile. The error text is actually useful, though:

a.cpp a.cpp(11) : error C3212: 'A::B' : an explicit specialization of a template member must be a member of an explicit specialization a.cpp(8) : see declaration of 'A::B'

It looks like it is only legal to fully specialize B if you also fully specialize A.

Edit: Ok, I heard back from someone who can speak authoritatively on this - to paraphrase, this is a very murky area in the standard, and it's an open issue with the C++ Committee to clean it up ("it" being explicit specializations of members of class templates). In the near term, the advice is "Don't do that".