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

[Lawrence Bottorff]

Why can I not do this?

data NestedList1 a = Elem a | NList (NestedList1 a)

that is, with parens rather than square brackets, then

myList2 = (NList (Elem 1, NList (Elem 2, NList (Elem 3, Elem 4), Elem 5)))

It gives the error

<interactive>:383:18-73: error:
    ,* Couldn't match expected type `NestedList1 a'
                  with actual type `(NestedList1 Integer, NestedList1 a0)'
    ,* In the first argument of `NList', namely
        `(Elem 1, NList (Elem 2, NList (Elem 3, Elem 4), Elem 5))'
      In the expression:
        (NList (Elem 1, NList (Elem 2, NList (Elem 3, Elem 4), Elem 5)))
      In an equation for `myList2':
          myList2
            = (NList (Elem 1, NList (Elem 2, NList (Elem 3, Elem 4), Elem
5)))
    ,* Relevant bindings include
        myList2 :: NestedList1 a (bound at <interactive>:383:1)

etc., etc.

On Tue, Jan 26, 2021 at 4:05 PM David McBride <toad3k at gmail.com> wrote:

> 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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