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

[Dmitriy Matrosov]

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.

it will hang. I don't understand, why it tries to compute second list 
(z2), if it only needs first?