Is it possible to "store" a template parameter pack without expanding it?

Question

I was experimenting with C++0x variadic templates when I stumbled upon this issue:

template < typename ...Args > struct identities {     typedef Args type; //compile error: "parameter packs not expanded with '...' };  //The following code just shows an example of potential use, but has no relation //with what I am actually trying to achieve. template < typename T > struct convert_in_tuple {     typedef std::tuple< typename T::type... > type; };  typedef convert_in_tuple< identities< int, float > >::type int_float_tuple; 

GCC 4.5.0 gives me an error when I try to typedef the template parameters pack.

Basically, I would like to "store" the parameters pack in a typedef, without unpacking it. Is it possible? If not, is there some reason why this is not allowed?

Answer 1

Another approach, which is slightly more generic than Ben's, is as follows:

#include <tuple>  template <typename... Args> struct variadic_typedef {     // this single type represents a collection of types,     // as the template arguments it took to define it };  template <typename... Args> struct convert_in_tuple {     // base case, nothing special,     // just use the arguments directly     // however they need to be used     typedef std::tuple<Args...> type; };  template <typename... Args> struct convert_in_tuple<variadic_typedef<Args...>> {     // expand the variadic_typedef back into     // its arguments, via specialization     // (doesn't rely on functionality to be provided     // by the variadic_typedef struct itself, generic)     typedef typename convert_in_tuple<Args...>::type type; };  typedef variadic_typedef<int, float> myTypes; typedef convert_in_tuple<myTypes>::type int_float_tuple;  int main() {}