# 99 questions/Solutions/17

### From HaskellWiki

(*) Split a list into two parts; the length of the first part is given.

Do not use any predefined predicates.

Solution usingtake

drop

split xs n = (take n xs, drop n xs)

splitAt

split = flip splitAt

But these should clearly be considered "predefined predicates". Alternatively, we have the following recursive solution:

split :: [a] -> Int -> ([a], [a]) split [] _ = ([], []) split l@(x : xs) n | n > 0 = (x : ys, zs) | otherwise = ([], l) where (ys,zs) = split xs (n - 1)

The same solution as above written more cleanly:

split :: [a] -> Int -> ([a], [a]) split xs 0 = ([], xs) split (x:xs) n = let (f,l) = split xs (n-1) in (x : f, l)