Deducing a function pointer return type

Question

I think code will better illustrate my need:

template 
struct return_type
{
  typedef ??? type;
};

so that:

Answer 1

If you can use variadic templates (November '12 CTP), this should work:

template <class F>
struct return_type;

template <class R, class... A>
struct return_type<R (*)(A...)>
{
  typedef R type;
};

Live example.

If you can't use variadic templates, you'll have to provide specific specialisations for 0, 1, 2, ... parameters (by hand or preprocessor-generated).

EDIT

As pointed out in the comments, if you want to work with variadic functions as well, you'll have to add one extra partial specialisation (or one for each parameter count in the no-variadic-templates case):

template <class R, class... A>
struct return_type<R (*)(A..., ...)>
{
  typedef R type;
};