[Haskell-cafe] Dynamically stackable monads

5 May 2006

      Hello,

I was wondering if it's possible to stack a runtime-known amount of
monads on top of each other. Let me illustrate. Assume I have a monad
that can consume data and expects as starting parameter an action of the
underlying monad to use this data (call it produce at the lower level
monad).

Now one could imagine stacking one of these consumers on top of the
other, as can be seen below. However I can not choose at runtime how
many I want to stack. Is there any solution for this?

Regards,
Christophe

------------------------------
--------------------------------------------------
{-# OPTIONS_GHC -fglasgow-exts #-}
module Main where
import Control.Monad
import Control.Monad.Identity
import Control.Monad.State
import Control.Monad.Trans

data Action e m = Action {
 produce :: e -> m ()
}

newtype SequencerT e m a = SequencerT (StateT (Action e m) m a)
 deriving (Functor, Monad, MonadIO)

newtype Sequencer e a = Sequencer (SequencerT e Identity a)
 deriving (Functor, Monad, MonadSequencer e)

instance MonadTrans (SequencerT e) where
 lift = SequencerT . lift

class Monad m => MonadSequencer e m | m -> e where
 consume     :: e -> m ()

instance Monad m => MonadSequencer e (SequencerT e m) where
 consume x   = SequencerT $ do
                 s <- get
                 lift . (produce s) $ x

evalSequencerT (SequencerT s) action =
 evalStateT s action
evalSequencer (Sequencer s) inputs action =
 evalSequencerT s action

runSequencerT (SequencerT s) action =
 runStateT s action
runSequencer (Sequencer s) action =
 runSequencerT s action

main :: IO () =
 evalSequencerT
   (evalSequencerT
     (consume 1 >> consume 2 >> consume 3)
     (Action{produce = \x -> if x > 1 then consume x else liftIO . print $
("A" ++ show x)}))
   (Action{produce = print . ("B" ++) . show })

--
Christophe Poucet
Ph.D. Student
Phone:+32 16 28 87 20
E-mail: --- <Christophe.Poucet@imec.be>
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