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Today's Topics:
1. Re: Defined list data type compared to Haskell List
(David McBride)
2. As pattern, @ (Lawrence Bottorff)
3. Re: As pattern, @ (Ut Primum)
----------------------------------------------------------------------
Message: 1
Date: Tue, 26 Jan 2021 18:14:27 -0500
From: David McBride <toa...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <Beginners@haskell.org>
Subject: Re: [Haskell-beginners] Defined list data type compared to
Haskell List
Message-ID:
<CAN+Tr432iej=QdpNKRgXgobe45=n88gauq55locsrqtkoml...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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 <borg...@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 <toa...@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 <borg...@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 <toa...@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 <borg...@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
>>>>> _______________________________________________
>>>>> Beginners mailing list
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>>>>>
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Message: 2
Date: Wed, 27 Jan 2021 00:01:23 -0600
From: Lawrence Bottorff <borg...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: [Haskell-beginners] As pattern, @
Message-ID:
<cafahfsxnzndskgppg-__cvghnohp2ucljrspzg-1gmgsryc...@mail.gmail.com>
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I'm looking at this <https://wiki.haskell.org/99_questions/Solutions/8> and
wondering how the "as pattern" actually is working
compress (x:ys@(y:_))
| x == y = compress ys
| otherwise = x : compress ys
compress ys = ys
I'm sure it's just some version of my stab at eliminating consecutive
duplicates in a list
compress :: Eq a => [a] -> [a]
compress [] = []
compress [x] = [x]
compress (x:y:xs) = if x == y then compress (y:xs) else x : compress (y:xs)
only smarter. Could someone walk me through the (x:ys@(y:_)) part?
LB
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Message: 3
Date: Wed, 27 Jan 2021 07:14:35 +0100
From: Ut Primum <utpri...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] As pattern, @
Message-ID:
<CANjDmKKsFvgKLgLVz4n4W-_t_Edga+TnWh-bqbm9O=vanil...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
Hi,
between the arguments of a function,
ys@(y:yss)
would mean that we call that argument ys, we call its head y and its tail
yss. In your case the tail is replaced by _ , because is never used inside
the function.
Il mer 27 gen 2021, 07:02 Lawrence Bottorff <borg...@gmail.com> ha scritto:
> I'm looking at this <https://wiki.haskell.org/99_questions/Solutions/8>
> and wondering how the "as pattern" actually is working
>
> compress (x:ys@(y:_))
> | x == y = compress ys
> | otherwise = x : compress ys
> compress ys = ys
>
> I'm sure it's just some version of my stab at eliminating consecutive
> duplicates in a list
>
> compress :: Eq a => [a] -> [a]
> compress [] = []
> compress [x] = [x]
> compress (x:y:xs) = if x == y then compress (y:xs) else x : compress (y:xs)
>
> only smarter. Could someone walk me through the (x:ys@(y:_)) part?
>
> LB
>
>
> _______________________________________________
> Beginners mailing list
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