Re: [Haskell-cafe] Interpreter with Cont

2011/11/21 David Menendez

Here's how you might implement your monad using Cont,

type InteractionM a b = Cont (Interaction a b)

exit b = Cont $ \k -> Exit b output b = Cont $ \k -> Output b (k ()) input = Cont $ \k -> Input k runM m = runCont m Exit

That's what I originally wanted to know. I guess I struggled with the definition of output. Oh, there's so much more to learn... Thanks, Tim