John Moeller writes:

Hello all,

I am trying to make an "is_range" metafunction that (essentially) returns mpl::true_ when the template parameter is a range (as defined by the boost::range library).

I can do this if I create a specialization for every "non-range" type currently specialized by the library (such as built-in arrays, char strings, etc.). However, I'd like to have is_range work for a user-defined range without the user *also* needing to specialize is_range.

I tried something like this:

////////////////////////////////////////////////////////////////////// BOOST_MPL_HAS_XXX_TRAIT_DEF(type)

template < bool B > struct is_range_impl : boost::mpl::false_ {};

template <> struct is_range_impl< true > : boost::mpl::true_ {};

template < typename R > struct is_range : is_range_impl< has_type< boost::range_iterator< R > >::value > {}; //////////////////////////////////////////////////////////////////////

Which does fine if R is a range (even when you use the non-intrinsic method of defining a custom class as a range), but if R is *not* a range, the compiler complains that "iterator" is not a member of the class that I'm passing as the parameter, because it defaults to using the embedded "iterator" type in the parameter.

Is there a better way to define is_range so that it will work? I imagine that I'd have to use SFINAE somehow, but I can't quite see how to put it together.

Many thanks ahead of time,

--

John Moeller

I figured out how to do it (and it's pretty simple): template < typename R, typename V = void > struct is_range_impl : boost::mpl::false_ {}; template < typename R > struct is_range_impl< R, boost::range_iterator< R > > : boost::mpl::true_ {}; template < typename R > struct is_range : is_range_impl< R > {}; Feel free to pick this apart; I wouldn't mind if it were made better. Thanks, -- John Moeller