[Haskell-cafe] ismzero operator possible without equal constraint

[edgar klerks]

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 <dave at zednenem.com> wrote:

> On Sat, Dec 3, 2011 at 3:55 PM, Antoine Latter <aslatter at gmail.com> wrote:
> > On Sat, Dec 3, 2011 at 10:55 AM, edgar klerks <edgar.klerks at gmail.com>
> 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 <dave at zednenem.com>
> <http://www.eyrie.org/~zednenem/>
>
-------------- next part --------------
An HTML attachment was scrubbed...
URL: <http://www.haskell.org/pipermail/haskell-cafe/attachments/20111204/1801315c/attachment.htm>