No not for lists, but it is not a bad direction. If I modify it a bit, I can get an ifmzero function: ifmzero :: (MonadSplit m) => m a -> m b -> m b -> m b ifmzero p b f = join $ mhead $ (liftM (const f) p) `mplus` (return b) mhead :: (MonadSplit m) => m a -> m a mhead = liftM fst . msplit Which I think works for all MonadSplit monads. I have some loose rationing, I can show, but am a bit affraid to share :) I have made a small example with a foldl function. Thanks for your example. Greets, Edgar

module Control.Monad.MonadSplit where import Control.Monad import Control.Applicative import qualified Data.Sequence as S import Test.QuickCheck

class MonadPlus m => MonadSplit m where msplit :: m a -> m (a, m a)

instance MonadSplit [] where msplit [] = mzero msplit (x:xs) = return (x,xs)

instance MonadSplit Maybe where msplit Nothing = mzero msplit (Just x) = return (x, Nothing)

ifmzero p b f = join $ mhead $ (liftM (const f) p) `mplus` (return b)

mhead :: (MonadSplit m) => m a -> m a mhead = liftM fst . msplit

foldMSl :: (MonadSplit m) => (b -> a -> m b) -> b -> m a -> m b foldMSl m i n = ifmzero n (return i) $ do (x, xs) <- msplit n i' <- m i x foldMSl m i' xs

prop_foldMSl_ref = property $ test_foldMSl_ref where test_foldMSl_ref :: Int -> [Int] -> Bool test_foldMSl_ref x y = (foldMSl (\x y -> return $ x - y) x y) == (return (foldl (\x y -> x - y) x y))

On Sat, Dec 3, 2011 at 11:39 PM, David Menendez wrote:

On Sat, Dec 3, 2011 at 3:55 PM, Antoine Latter wrote:

...

On Sat, Dec 3, 2011 at 10:55 AM, edgar klerks wrote:

...

Hi list,

I am using MonadSplit (from http://www.haskell.org/haskellwiki/New_monads/MonadSplit ) for a project and now I want to make a library out of it. This seems to be straightforward, but I got stuck when I tried to move miszero out of the class:

miszero :: m a -> Bool

It tests if the provided monad instance is empty. My naive attempt was:

You can write:

miszero :: MonadPlus m => m a -> m Bool miszero m = (m >> return False) <|> return True

but that will invoke any monadic effects as well as determining the nature of the value, which may not be what you want.

It's almost certainly not what you want for the list monad.

-- Dave Menendez http://www.eyrie.org/~zednenem/