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

c++, c++11, templates, variadic-templates

Solution

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()
{}

Problem

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?

Original source