hi, you might find the "backward" state monad interesting. here is the basic idea: newtype S s a = S (s -> (a,s)) instance Monad (S s) where return a = S (\s -> (a,s)) S m >>= k = S (\s1 -> let (a,s3) = m s2 (b,s2) = run s1 (k a) in (b,s3)) put x = S (\s -> ((),x:s)) run s (S m) = m s test = snd $ run [] $ do put 'x' put 'y' undefined hope this helps -iavor On 12/30/06, Ryan Ingram <ryani.spam@gmail.com> wrote:
Hi everyone... it's my newbie post!
I am trying to create a monad which allows computations to output data to a stream. (Probably such a thing already exists, but it's a good problem for my current skill level in Haskell)
For example:
streamDemo = do output 1 output 2 output 5
makelist streamDemo -- [1,2,5]
I modelled my implementation around the state monad, but with a different execution model:
class (Monad m) => MonadStream w m | m -> w where output :: w -> m () run :: m a -> s -> (s -> w -> s) -> s -- basically foldl on the stream values makelist m = reverse $ run m [] (flip (:))
-- s is the type of the object to stream, r is the return type type StreamFunc s r = forall b. b -> (b -> s -> b) -> (r,b) newtype Stream s r = Stream { run' :: StreamFunc s r } instance Monad (Stream s) where return r = Stream (\s _ -> (r,s)) Stream m >>= k = Stream (\s f -> let (r,s') = (m s f) in run' (k r) s' f) instance (MonadStream w) (Stream w) where output w = Stream (\s f -> ((),f s w)) run m st f = snd $ run' m st f
What I don't like is how makelist comes out. It feels wrong to need to use reverse, and that also means that infinite streams completely fail to work. But I think it's impossible to fix with the "foldl"-style "run". Is there a better implementation of "makelist" possible with my current definition of "run"? If not, what type should "run" have so that it can work correctly?
As an example, I want to fix the implementation to make the following code work: fibs :: Stream Integer () fibs = fibs' 0 1 where fibs' x y = output y >> fibs' y (x+y)
fiblist :: [Integer] fiblist = makelist fibs
take 5 fiblist -- [1,1,2,3,5], but currently goes into an infinite loop
Thanks, -- ryan _______________________________________________ Haskell mailing list Haskell@haskell.org http://www.haskell.org/mailman/listinfo/haskell