On Sun, Oct 31, 2004 at 06:37:20PM +0100, Lemming wrote:
I encountered that the implementation of 'partition' in GHC 6.2.1 fails on infinite lists:
partition :: (a -> Bool) -> [a] -> ([a],[a]) partition p xs = foldr (select p) ([],[]) xs
select p x (ts,fs) | p x = (x:ts,fs) | otherwise = (ts, x:fs)
Ah, IIRC one of my very first haskell-posts was about this :) Actually, AFAICS this isn't just a could-be-better, but a real Bug(TM). According to The Report the definition is: partition p xs = (filter p xs, filter (not . p) xs) which doesn't have any trouble with infinite lists.
With the following definition we don't have this problem:
partition :: (a -> Bool) -> [a] -> ([a], [a]) partition _ [] = ([],[]) partition p (x:xs) = let (y,z) = partition p xs in if p x then (x : y, z) else (y, x : z)
Cheers, Remi -- Nobody can be exactly like me. Even I have trouble doing it.