How do I determine if a type is callable with only const references?

Question

I want to write a C++ metafunction is_callable that defines value to be true, if and only if the type F has the function cal

Answer 1

Ran across this while doing something else, was able to adapt my code to fit. It has the same features (and limitations) as @Xeo, but does not need sizeof trick/enable_if. The default parameter takes the place of needing to do the enable_if to handle template functions. I tested it under g++ 4.7 and clang 3.2 using the same test code Xeo wrote up

#include 
#include 

namespace detail {
  template
  struct call_exact : std::false_type {};

  template struct ARGS { typedef void type; };

  template
  C * opclass(decltype(std::declval()(std::declval()...)) (C::*)(Args...)) { }
  template
  C * opclass(decltype(std::declval()(std::declval()...)) (C::*)(Args...) const) { }

  template
  struct call_exact,
    typename ARGS<
       decltype(std::declval()(std::declval()...)),
       decltype(opclass(&T::operator()))
     >::type
   > : std::true_type {};
}

template
struct Callable : detail::call_exact> { };

template
struct Callable
 : Callable, Args...>{};