Subject: Re: [Boost-users] fusion transform_view
From: Stefan Strasser (strasser_at_[hidden])
Date: 2010-09-12 11:31:24
Zitat von Hicham Mouline <hicham_at_[hidden]>:
> result_of::transform<Seq, UnaryFct>
> metafunction, the result type I get, printed out by typeid(...).name() gives
>
> transform_view< Seq, UnaryFct, void_ >
>
> is there a way to print out the actual resulting type, forcing the view to
> "evaluate" ?