hi,
the foldr definition can be fixed by putting a ~ on the pattern in 'select'.
-iavor
Remi Turk wrote:
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
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