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

26 Jan 2021

      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  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  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 
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  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 
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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