Data.List.partition on infinite lists
Remi Turk
rturk at science.uva.nl
Sun Oct 31 20:23:35 EST 2004
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.
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