[Haskell] Converting a 'streaming' monad into a list

Ryan Ingram ryani.spam at gmail.com
Sat Dec 30 06:21:01 EST 2006

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
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
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

  -- ryan
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