On Mon, Apr 16, 2012 at 12:52:02PM +0400, Dmitriy Matrosov wrote:
Hi all.
If i implement take using foldr
take' :: Int -> [a] -> [a] take' n = foldr (\x z -> if fst x <= n then snd x : z else []) [] . zip [1..]
it'll work fine with infinite lists. But if i implement splitAt similarly
splitAt' :: Int -> [a] -> ([a], [a]) splitAt' n = foldr (\x (z1, z2) -> if fst x <= n then (snd x : z1, z2) else ([], snd x : z2)) ([], []) . zip [1..]
and call it like this
*Main> fst $ splitAt' 4 [1..] ^CInterrupted.
Try something like this: splitAt' n = foldr (\x zs -> if fst x <= n then (snd x : fst zs, snd zs) else ([], snd x : snd zs)) ([], []) . zip [1..] I'm no Haskell expert, but I suspect that when pattern-matching z2, it tries to evaluate it and it hangs... My version does not hang... hth, L. -- Lorenzo Bolla http://lbolla.info