# 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 using take and drop:

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

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)
```

Note that this function, with the parameters in the other order, exists as `splitAt`

.