From: "Joel de Guzman" <[EMAIL PROTECTED]> [...] > The condition is expected to be a functor that returns > a boolean condition. I was hoping that I can use the > ref(b) as a functor such that I can write: > > bool b; > > if_p(ref(b)) > [ > parse_this > ]
Not a good idea. In lambda terms the above is var(b), not ref(b). We've been using ref(x) to mean "just like x", i.e. ref(b)() means b(). One day we might even get that core change that would enable reference_wrapper<T>::operator T& to be considered in a ref(f)() expression; for now, we "fake" it in bind, function, etc. _______________________________________________ Unsubscribe & other changes: http://lists.boost.org/mailman/listinfo.cgi/boost