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

[David McBride]

You can, you have to write it slightly differently. But defined that way it
can only ever have one element because NList only takes one.

mylist = NList (NList (NList (Elem 1)))

You could also change it to

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

But at that point you have a nonempty list type (always at least one
element) which already exists.

On Tue, Jan 26, 2021, 17:40 Lawrence Bottorff <borgauf at gmail.com> wrote:

> 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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