Can one access the template parameter outside of a template without a typedef?

Question

A simple example:

template<typename _X> // this template parameter should be usable outside!
struct Small {
   typedef _X X; // this is tedious!
   X foo;
};

template<typename SomeSmall>
struct Big {
   typedef typename SomeSmall::X X; // want to use X here!
   SomeSmall bar;
   X toe;
};

Is there a way to access the template parameter X of Small without using a typedef in the Small class?

Answer 1

Yes, define a second "getter" template with partial specialization.

template< typename >
struct get_Small_X; // base template is incomplete, invalid

template< typename X > // only specializations exist
struct get_Small_X< Small< X > > {
    typedef X type;
};

Now instead of Small<X>::X you have typename get_Small_X< Small<X> >::type.

By the way, _X is a reserved identifier, so you shouldn't use it for anything. X_ is a better choice.

---

Advanced topic: template introspection.

While I'm thinking about it, you don't need to define this separately for every template. A single master template should do it.

This compiles in Comeau, I know there are rules about matching template template arguments but I think it's OK… template template arguments are forbidden from the master template in partial specialization.

template< typename >
struct get_first_type_argument;

template< template< typename > class T, typename X >
struct get_first_type_argument< T< X > > {
    typedef X type;
};

template< typename X >
struct simple;

get_first_type_argument< simple< int > >::type q = 5;

This only works with "unary" templates but could be adapted in C++0x for the general case.