Re: [Haskell-cafe] Re: Can we come out of a monad?

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