Variadic可变参数模板模板

Question

我目前正在努力使用以下代码,其目的是实现可变参数可变参数模板模板:

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;
};

理想情况下,我可以使用此模板元函数来实现以下功能:

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;
*/

上述的输出有望创建new_obj1和new_obj2形式template<int, int, double, double, char, char*, int, double, const char*> struct new_obj[1|2] {};

我在Windows上使用gcc 4.6.2,它为" Head<typename TailArgs...>..." 的扩展输出了"'...之前的预期参数包' ".

使用gcc 4.5.1可以重现此错误.

Answer 1

尝试这样的事情:

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;
};