How can I have multiple parameter packs in a variadic template?

Question

Function one() accepts one parameter pack. Function two() accepts two. Each pack is constrained to be wrapped in types A and B. Why is it

Answer 1

Here is another way to have several parameters packs using template template parameters:

#include <iostream>

template <typename... Types>
struct foo {};

template < typename... Types1, template <typename...> class T
         , typename... Types2, template <typename...> class V
         , typename U >
void
bar(const T<Types1...>&, const V<Types2...>&, const U& u)
{
  std::cout << sizeof...(Types1) << std::endl;
  std::cout << sizeof...(Types2) << std::endl;
  std::cout << u << std::endl;
}

int
main()
{
  foo<char, int, float> f1;
  foo<char, int> f2;
  bar(f1, f2, 9);
  return 0;
}

Answer 2

Function templates (like skypjack's example) and partial specializations of class and variable templates can have multiple parameter packs if each template parameter subsequent to a template parameter pack either has a default value or can be deduced. The only thing I'd like to add/point out is that for class and variable templates you need a partial specialization. (See: C++ Templates, The Complete Guide, Vandevoorde, Josuttis, Gregor 12.2.4, Second Edition)

// A template to hold a parameter pack
template < typename... >
struct Typelist {};

// Declaration of a template
template< typename TypeListOne 
        , typename TypeListTwo
        > 
struct SomeStruct;

// Specialization of template with multiple parameter packs
template< typename... TypesOne 
        , typename... TypesTwo
        >
struct SomeStruct< Typelist < TypesOne... >
                 , Typelist < TypesTwo... >
                 >
{
        // Can use TypesOne... and TypesTwo... how ever
        // you want here. For example:
        typedef std::tuple< TypesOne... > TupleTypeOne;
        typedef std::tuple< TypesTwo... > TupleTypeTwo;
};      

Answer 3

The compiler needs a way to know where is the barrier between the two variadic templates. A clean way of doing this is to define one pack of arguments for an object and the second pack for a static member function. This can be appied to more than two variadic templates by nesting multiple structs in eachother. (keeping the last level as a function)

#include <iostream>

template<typename... First>
struct Obj
{
    template<typename... Second>
    static void Func()
    {
        std::cout << sizeof...(First) << std::endl;
        std::cout << sizeof...(Second) << std::endl;
    }
};

int main()
{
    Obj<char, char>::Func<char, char, char, char>();
    return 0;
}

Answer 4

I found one solution. Wrap each parameter pack in a Tuple. Use a struct for partial specialization. Here's a demo that forwards arguments to a functor by consuming one tuple as a list and accumulating another. Well, this one forwards by copying. Tuples are used in type deduction yet no tuples are used in function parameters, which I think is neat.

#include <iostream>
#include <tuple>

template < typename ... >
struct two_impl {};

// Base case
template < typename F,
           typename ...Bs >
struct two_impl < F, std::tuple <>, std::tuple< Bs... > >  {
  void operator()(F f, Bs... bs) {
    f(bs...);
  }
};

// Recursive case
template < typename F,
           typename A,
           typename ...As,
           typename ...Bs >
struct two_impl < F, std::tuple< A, As... >, std::tuple< Bs...> >  {
  void operator()(F f, A a, As... as, Bs... bs) {
    auto impl = two_impl < F, std::tuple < As... >, std::tuple < Bs..., A> >();
    impl(f, as..., bs..., a);
  }
};

template < typename F, typename ...Ts >
void two(F f, Ts ...ts) {
  auto impl = two_impl< F, std::tuple < Ts... >, std::tuple <> >();
  impl(f, ts...);
}

struct Test {
  void operator()(int i, float f, double d) {
    std::cout << i << std::endl << f << std::endl << d << std::endl;
  }
};

int main () {
  two(Test(), 1, 1.5f, 2.1);
}

Tuples are a very good compile time list.