On 9/8/2012 12:45 PM, Michel Morin wrote:
template < class , class = void > struct result_of_aux {};
template < class F, class ...Args> struct result_of_aux<F(Args...), decltype(std::declval<F>()(std::declval<Args>()...), void ())> { typedef decltype(std::declval<F>()(std::declval<Args>()...)) type; };
What is void()? That's not any expression I recognize. It's a function type. Anyway, keep in mind that std::declval<F>()(...) could have an incomplete type, so you can't really use it in an expression. You should use it in decltype, and then pass it as a template parameter to something like this: template<typename T> struct always_void { typedef void type; }; template < class , class = void > struct result_of_aux {}; template < class F, class ...Args> struct result_of_aux<F(Args...), typename always_void<decltype(std::declval<F>()(std::declval<Args>()...))>::type> { typedef decltype(std::declval<F>()(std::declval<Args>()...)) type; }; HTH, -- Eric Niebler BoostPro Computing http://www.boostpro.com