Implement `flatten` for `Cons`

Question

Given the following definition of Cons:

data Cons a = Cons a (Cons a) 
             | Empty 
             deriving Show

I implemented a flatten function.

It's mean to work exactly like concat :: [[a]] -> [a], but for Cons.

flatten :: Cons (Cons a) -> Cons a
flatten Empty                = Empty
flatten (Cons (Empty) ys)    = flatten ys
flatten (Cons (Cons x xs) ys) = Cons x (flatten (Cons xs ys))

Test data:

test1 :: Cons (Cons Int)
test1 = Cons (Cons 5 Empty) Empty

test2 :: Cons (Cons Int)
test2 = Cons (Cons 5 (Cons 10 Empty)) Empty

test3 :: Cons (Cons Int)
test3 = Cons (Cons 5 (Cons 10 (Cons 20 Empty))) test2

Some tests:

ghci> flatten test1
Cons 5 Empty

ghci> flatten test2
Cons 5 (Cons 10 Empty)

ghci> flatten test3
Cons 5 (Cons 10 (Cons 20 (Cons 5 (Cons 10 Empty))))

Answer 1

Well, unless you wanted to add instances to allow for folding and so, or conversions to Data.List I'd say that's perfect as it is.