Why doesn't this function work if I use “[xs]” instead of “xs”?

Question

split :: [a] -> Int -> ([a], [a])
split [xs] n = 
    (take n [xs], drop n [xs])

The same code works if I give the variable as xs

Answer 1

Somehow a lot of people think that [xs] as pattern means that you unify a list with xs. But this is incorrect, since the function signature (either derived implicitly, or stated explicitly) already will prevent you to write code where you call the function with a non-list item.

A list has two constructors:

• the empty list []; and
• the "cons" (h : t) with h the head (first element), and t the tail (a list with the remaining elements).

Haskell however introduces some syntactical sugar as well. For example [1] is short for (1:[]), and [1, 4, 2] for (1:(4:(2:[]))).

So that means that if you write [xs], behind the curtains you defined a pattern (xs: []) which thus means you match all lists with exactly one element, and that single element (not the entire list) is then xs.

Anyway, the solution is to use:

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

Since both take :: Int -> [a] -> [a] and drop :: Int -> [a] -> [a] have in the signature that xs is supposed to be a list, Haskell will derive automatically that n is supposed to be an Int, and xs an [a].

Note that you can use splitAt :: Int -> [a] -> ([a], [a]) as well. We can make the signature equivalent to the one you target with:

split = flip splitAt