¿Es posible “tienda” un paquete de parámetro de plantilla sin expandiéndolo?

Question

Estaba experimentando con C ++ 0x variadic plantillas cuando me encontré con este tema:

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 me da un error cuando trato de typedef el paquete de plantillas parámetros.

Básicamente, me gustaría "almacenar" los parámetros de empacar en un typedef, sin sacarlo del embalaje. ¿Es posible? Si no, ¿hay alguna razón por la que esto no está permitido?

Answer 1

Otro enfoque, que es ligeramente más genérico que Ben, es el siguiente:

#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()
{}

Answer 2

Creo que la razón por la que no está permitido es que sería complicado, y se puede trabajar alrededor de ella. Es necesario utilizar la inversión de dependencia y hacer que la estructura de almacenar el paquete de parámetros en una plantilla de la fábrica capaz de aplicar ese paquete parámetro a otra plantilla.

Algo a lo largo de las líneas de:

template < typename ...Args >
struct identities
{
    template < template<typename ...> class T >
    struct apply
    {
        typedef T<Args...> type;
    };
};

template < template<template<typename ...> class> class T >
struct convert_in_tuple
{
    typedef typename T<std::tuple>::type type;
};

typedef convert_in_tuple< identities< int, float >::apply >::type int_float_tuple;

Answer 3

I've found Ben Voigt's idea very useful in my own endeavors. I've modified it slightly to make it general to not just tuples. To the readers here it might be an obvious modification, but it may be worth showing:

template <template <class ... Args> class T, class ... Args>
struct TypeWithList
{
  typedef T<Args...> type;
};

template <template <class ... Args> class T, class ... Args>
struct TypeWithList<T, VariadicTypedef<Args...>>
{
  typedef typename TypeWithList<T, Args...>::type type;
};

The name TypeWithList stems from the fact that the type is now instantiated with a previous list.

Answer 4

This is a variation of GManNickG's neat partial specialization trick. No delegation, and you get more type safety by requiring the use of your variadic_typedef struct.

#include <tuple>

template<typename... Args>
struct variadic_typedef {};

template<typename... Args>
struct convert_in_tuple {
    //Leaving this empty will cause the compiler
    //to complain if you try to access a "type" member.
    //You may also be able to do something like:
    //static_assert(std::is_same<>::value, "blah")
    //if you know something about the types.
};

template<typename... Args>
struct convert_in_tuple< variadic_typedef<Args...> > {
    //use Args normally
    typedef std::tuple<Args...> type;
};

typedef variadic_typedef<int, float> myTypes;
typedef convert_in_tuple<myTypes>::type int_float_tuple; //compiles
//typedef convert_in_tuple<int, float>::type int_float_tuple; //doesn't compile

int main() {}