boost::is_function VS compile error (for const operator function)

[Yaroslav Lukyanchuk <yarosladov@…>]

If you try to evaluate boost::is_function on functions that have the same signature and one of them are member const operator function, VC++ (10 through 12) fires compile error:

Sample minimal code:

struct Functor {

void operator()() const {} };

boost::is_function<void()>::value; boost::is_function<decltype(Functor().operator())>::value;

Compiler output: 1>$(path)\boost\type_traits\is_function.hpp(72): error C2373: 't' : redefinition; different type modifiers 1> $(path)\boost\type_traits\is_function.hpp(72) : see declaration of 't' 1> $(path)\boost\type_traits\is_function.hpp(97) : see reference to class template instantiation 'boost::detail::is_function_impl<T>' being compiled

[Yaroslav Lukyanchuk <yarosladov@…>]

sample code

[John Maddock]

I'm not completely sure but I think there's a problem with your use of decltype: gcc for example won't allow it to be used in this way (even if you remove the is_function usage). If you pass the actual type of the member function to is_function then there's no issue. So my suspicion is that is_function is a bystander in this.

[John Maddock]

Double checking, I'm pretty sure the code is mal-formed, the correct form is:

boost::is_function<decltype(std::declval<Functor>()())>::value;

Which does work just fine in your example.