So NestedList is using regular List? So in

data NestedList a = Elem a | List [NestedList a]

the second data constructor List [NestedList a] we see a "regular" list because of the square brackets?

On Tue, Jan 26, 2021 at 3:42 PM David McBride <toad3k@gmail.com> wrote:

In NestedList, the List constructor takes a regular list of NestedLists.  Therefore when pattern matching on it you can get access to those nested lists.  In your code, x is the first NestedList, and xs is the rest of the NestedLists.

On Tue, Jan 26, 2021 at 4:32 PM Lawrence Bottorff <borgauf@gmail.com> wrote:

I'm following this and yet I see this solution

data NestedList a = Elem a | List [NestedList a] deriving (Show)

flatten1 :: NestedList a -> [a]
flatten1 (Elem a   )   = [a]
flatten1 (List (x:xs)) = flatten1 x ++ flatten1 (List xs)
flatten1 (List [])     = []

What I find puzzling is this line

flatten1 (List (x:xs)) = flatten1 x ++ flatten1 (List xs)

where I see 

(List (x:xs)) as an argument. How is the NestedList type also able to be expressed as a normal consed list with x:xs argument? How is (:) interacting with NestedList?

LB

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