Variadic variadic template templates

Question

I'm currently struggling with the following code, the intent of which is to implement variadic variadic template templates:

template
<
  template <typename... HeadArgs> class Head,
  template <typename... TailArgs> class...
>
struct join<Head<typename HeadArgs...>, Head<typename TailArgs...>...>
{
  typedef Head<typename HeadArgs..., typename TailArgs......> result;
};

Ideally, I'd be able to use this template metafunction to achieve the following:

template <typename...> struct obj1 {};
template <typename...> struct obj2 {};

typedef join
<
  obj1<int, int, double>, 
  obj1<double, char>,
  obj1<char*, int, double, const char*>
>::result new_obj1;

typedef join
<
  obj2<int, int, double>, 
  obj2<double, char>,
  obj2<char*, int, double, const char*>
>::result new_obj2;

/* This should result in an error, because there are 
   different encapsulating objects
typedef join
<
  obj1<int, int, double>, 
  obj1<double, char>,
  obj2<char*, int, double, const char*>
>::result new_obj;
*/

The output of the above would hopefully create new_obj1 and new_obj2 in the form template<int, int, double, double, char, char*, int, double, const char*> struct new_obj[1|2] {};

I'm using gcc 4.6.2 on Windows, which outputs an "expected parameter pack before '...'" for the expansion of "Head<typename TailArgs...>...".

This error is reproducable with gcc 4.5.1.

The argument name inside the template specification of the template template argument is optional and entirely cosmetic; it's not actually available as a real parameter. — Kerrek SB
– Kerrek SB, Commented Mar 12, 2012 at 6:33
@KerrekSB is there any way to get the parameters of a template template parameter? — kmore
– kmore, Commented Mar 12, 2012 at 6:43
I edited my post; I was missing the variadic case. In answer to your question: Yes, by supplying an explicit template parameter (pack) and matching. — Kerrek SB
– Kerrek SB, Commented Mar 12, 2012 at 6:53
@kmore No, because once you provide parameters to a template, it's no longer a template. template_name< parameters > is a class and will match a plain class or typename template parameter. — Potatoswatter
– Potatoswatter, Commented Mar 12, 2012 at 6:54
Indeed. You should notice that join isn't actually matching templates, but types. So the templates are just a detail of the specialization, not a defining characteristic of join. — Kerrek SB
– Kerrek SB, Commented Mar 12, 2012 at 6:59

Kerrek SB · Accepted Answer · 2012-03-12 06:53:07Z

25

Try something like this:

template <typename...> struct join;

template <template <typename...> class Tpl,
          typename ...Args1,
          typename ...Args2>
struct join<Tpl<Args1...>, Tpl<Args2...>>
{
    typedef Tpl<Args1..., Args2...> type;
};

template <template <typename...> class Tpl,
          typename ...Args1,
          typename ...Args2,
          typename ...Tail>
struct join<Tpl<Args1...>, Tpl<Args2...>, Tail...>
{
     typedef typename join<Tpl<Args1..., Args2...>, Tail...>::type type;
};

edited Mar 12, 2012 at 6:53

answered Mar 12, 2012 at 6:35

Kerrek SB

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1 Comment

kmore Over a year ago

That's a perfect solution when combining the parameters of two objects. However, I'm looking to combine the parameters of a variable number of objects. I tried modifying your answer, but there's still an error with the expansion pack. EDIT: you beat me to it, thank you!

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