On Thu, Oct 4, 2012 at 1:17 AM, Matt Calabrese <rivorus@gmail.com> wrote:
On Thu, Oct 4, 2012 at 4:13 AM, Matt Calabrese <rivorus@gmail.com> wrote:
Updated and tested for edge cases int GCC and Clang.
Updated one more time to simplify it a little. I'm going to sleep, otherwise I'll keep changing it.
Thanks a lot! template< class > struct always_void { typedef void type; }; template< class T, class U, class = void > struct common_type_impl_2 {}; // (1) template< class T, class U > struct common_type_impl_2 // (2) < T, U , typename always_void < decltype( true ? std::declval< T >() : std::declval< U >() ) >::type // (*) > { typedef decltype( true ? std::declval< T >() : std::declval< U >() ) type; }; So I understand it :) Would you explain me who the 2nd common_type_impl2 get selected when there's a common type given that always_void<...>::type is always void? Is it because when decltype( true ? std::declval< T >() : std::declval< U >() ) >::type is not valid in the instatiation of always_void (*) then SFINAE removes the overload (2) and (1) gets selected for being the only choice left? If so, why (1) has the lower priority in being selected when there's a common type and both (1) and (2) are valid? Thanks! --Lorenzo