[Haskell-beginners] Defined list data type compared to Haskell List

[David McBride]

Right that is a plain list of NestedLists.  So if you were to rewrite [a]
as (Regularlist a) so to speak (not a real type), the definition of
NestedList would be List (RegularList (NestedList a)).

Keep in mind that List is a constructor, everything after it is types.

On Tue, Jan 26, 2021 at 4:50 PM Lawrence Bottorff <borgauf at gmail.com> wrote:

> 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 at 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 at gmail.com>
>> wrote:
>>
>>> 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?
>>>
>>> LB
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