[Haskell-cafe] ismzero operator possible without equal constraint
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: miszero :: (Eq (m a), MonadPlus m) = m a - Bool miszero = ( == mzero ) This works, but not correctly. It adds an Eq constraint that is unneeded. I would prefer to have something like: miszero :: MonadPlus m = m a - Bool Because I am not comparing the contents of the monad. I don't even touch it. Is this possible to write? with kind regards, Edgar ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] ismzero operator possible without equal constraint
Of course it is not possible! Take a simple composition of reader and Maybe functors for an example: miszero :: (b - Maybe a) - Bool I'm pretty sure (b - Maybe a) for a is a MonadPlus, but you can't implement miszero for it. Arseniy. On 3 December 2011 16:55, edgar klerks edgar.kle...@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: miszero :: (Eq (m a), MonadPlus m) = m a - Bool miszero = ( == mzero ) This works, but not correctly. It adds an Eq constraint that is unneeded. I would prefer to have something like: miszero :: MonadPlus m = m a - Bool Because I am not comparing the contents of the monad. I don't even touch it. Is this possible to write? with kind regards, Edgar ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] ismzero operator possible without equal constraint
Hi Arseniy, Yes, I see it now. :) . I had some feeling there should be some structural equality: Just _ == Just _ = True Nothing == Nothing = True _ == _ = False But this doesn't work for functions. Thanks for your answer! Greets, Edgar On Sat, Dec 3, 2011 at 6:23 PM, Arseniy Alekseyev arseniy.alekse...@gmail.com wrote: Of course it is not possible! Take a simple composition of reader and Maybe functors for an example: miszero :: (b - Maybe a) - Bool I'm pretty sure (b - Maybe a) for a is a MonadPlus, but you can't implement miszero for it. Arseniy. On 3 December 2011 16:55, edgar klerks edgar.kle...@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: miszero :: (Eq (m a), MonadPlus m) = m a - Bool miszero = ( == mzero ) This works, but not correctly. It adds an Eq constraint that is unneeded. I would prefer to have something like: miszero :: MonadPlus m = m a - Bool Because I am not comparing the contents of the monad. I don't even touch it. Is this possible to write? with kind regards, Edgar ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] ismzero operator possible without equal constraint
On Sat, Dec 3, 2011 at 10:55 AM, edgar klerks edgar.kle...@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. Antoine miszero :: (Eq (m a), MonadPlus m) = m a - Bool miszero = ( == mzero ) This works, but not correctly. It adds an Eq constraint that is unneeded. I would prefer to have something like: miszero :: MonadPlus m = m a - Bool Because I am not comparing the contents of the monad. I don't even touch it. Is this possible to write? with kind regards, Edgar ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] ismzero operator possible without equal constraint
On Sat, Dec 3, 2011 at 3:55 PM, Antoine Latter aslat...@gmail.com wrote: On Sat, Dec 3, 2011 at 10:55 AM, edgar klerks edgar.kle...@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 d...@zednenem.com http://www.eyrie.org/~zednenem/ ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
Re: [Haskell-cafe] ismzero operator possible without equal constraint
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 d...@zednenem.com wrote: On Sat, Dec 3, 2011 at 3:55 PM, Antoine Latter aslat...@gmail.com wrote: On Sat, Dec 3, 2011 at 10:55 AM, edgar klerks edgar.kle...@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 d...@zednenem.com http://www.eyrie.org/~zednenem/ ___ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
