Excerpts from Ertugrul Soeylemez's message of Tue Aug 10 02:31:14 -0400 2010: > There is no evalCont, there is runCont: > > runCont :: (a -> r) -> Cont r a -> r > > Note that Cont/ContT computations result in a value of type 'r': > > newtype Cont r a = Cont ((a -> r) -> r)
Yes, but if you pass in 'id' as the continuation to 'runCont', the entire expression will result in 'a'. The continuation monad doesn't act globally... Still confused, Edward _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
