C++0x: variadic template puzzle

Hi!

Suppose I stored some parameters in a tuple so I can forward these
parameters _later_ to some function object. I just don't want to
forward the tuple itself but to excract the parameters from the tuple.
However, I have trouble figuring out how to do that. The function
templates I try to define are

template<typename Func, typename... Args>
void tuple_forward(Func && f, tuple<Args...> && args);

template<typename Func, typename... Args>
void tuple_forward(Func && f, tuple<Args...> const& args);

The first takes a tuple rvalue and should extract and forward all
parameters properly to f so it works for move-only types as well. The
second function simply passes the extracted parameters directly to f.
I came up with the following helper class:

   template<unsigned... Indices>
   struct invoke_helper
   {
      template<typename Func, typename... Args>
      static inline void doit(Func && f, tuple<Args...> && args)
      {
         f ( forward<Args>(get<Indices>(args))... );
      }
      template<typename Func, typename... Args>
      static inline void doit(Func && f, tuple<Args...> const& args)
      {
         f ( get<Indices>(args)... );
      }
   };

The problem is to generate the template parameters "Indices". Here is
an implementation attempt for the first function template:

   template<typename Func, typename... Args>
   inline void tuple_forward(Func && f, tuple<Args...> && args)
   {
      typedef invoke_helper<0,1,2,...,(sizeof...(Args)-1)> ih;
      ih::doit( forward<Func>(f) , move(args) );
      // Note: "args" is always an rvalue reference. There is
      // no reference collapsing involved.
   }

I havn't yet been able to come up with a solution for the "ih"-
typedef. Obviously the line is ill-formed. But it shows the intention.
Is there any solution to this problem? It yes, what does it look like?

Cheers!
SG

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