"Edward Z. Yang" <ezy...@mit.edu> wrote: > Excerpts from Luke Palmer's message of Tue Aug 10 01:04:04 -0400 2010: > > Except, of course, you want the signature > > > > evalCont :: Cont r a -> a > > > > Which is not possible. But I am not sure where all this discussion > > is coming from, Maybe and (r ->) cannot be broken out of. Isn't > > that example enough? > > I'm confused... that's the type of evalCont, no?
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) Greets, Ertugrul -- nightmare = unsafePerformIO (getWrongWife >>= sex) http://ertes.de/ _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
