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
I'm following this https://wiki.haskell.org/99_questions/Solutions/7 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?
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