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 <[EMAIL PROTECTED]> 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
[email protected]
http://www.haskell.org/mailman/listinfo/haskell
_______________________________________________
Haskell mailing list
[email protected]
http://www.haskell.org/mailman/listinfo/haskell
