You can use 'mapM' similar to 'map', to get the resultant list in the monad, and then liftM the function 'or' into it. This way you don't need to recurse explicitly, like in or' and and'. many :: Monad m => (a -> m Bool) -> [a] -> m Bool many p list = or `liftM` mapM p list (Type of mapM is: Monad m => (a -> m b) -> [a] -> m [b]) On Thu, Dec 29, 2011 at 7:45 AM, Manfred Lotz wrote:

Hi there, might be trivial but anyway.

I have a usage of 'any' and 'all' but I only have a predicate p :: a -> IO Bool.

I wrote my own functons for this:

mor :: Monad m => [m Bool] -> m Bool mor = liftM or . sequence

mand :: Monad m => [m Bool] -> m Bool mand = liftM and . sequence

or' :: Monad m => ( a -> m Bool) -> [a] -> [m Bool] or' _ [] = [] or' p (x:xs) = p x : or' p xs

and' :: Monad m => ( a -> m Bool) -> [a] -> [m Bool] and' _ [] = [] and' p (x:xs) = p x : and' p xs

myany :: Monad m => (a -> m Bool) -> [a] -> m Bool myany p = mor . or' p

myall :: Monad m => (a -> m Bool) -> [a] -> m Bool myall p = mand . and' p

which seems to do what I want.

Question: Is there any libray function I could use to do this?

-- Thanks, Manfred

_______________________________________________ Beginners mailing list Beginners@haskell.org http://www.haskell.org/mailman/listinfo/beginners

-- Markus Läll

On Thu, Dec 29, 2011 at 06:45:27AM +0100, Manfred Lotz wrote:

or' :: Monad m => ( a -> m Bool) -> [a] -> [m Bool] or' _ [] = [] or' p (x:xs) = p x : or' p xs

and' :: Monad m => ( a -> m Bool) -> [a] -> [m Bool] and' _ [] = [] and' p (x:xs) = p x : and' p xs

Note that or' = and' = map. -Brent