Re: [Haskell-beginners] splitAt implementation (using foldr) and infinite lists

On 04/16/12 18:21, Lorenzo Bolla wrote:

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.

Thanks, Lorenzo! It works now. On 04/16/12 18:55, Ozgur Akgun wrote:

You can also use lazy pattern matching.

http://en.wikibooks.org/wiki/Haskell/Laziness#Lazy_pattern_matching

On 16 April 2012 15:21, Lorenzo Bolla mailto:lbolla@gmail.com> wrote:

> 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..]

Ozgur Thanks, Ozgur!