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Today's Topics:
1. Re: Understanding functions like f a b c = c $ b a (Austin Zhu)
----------------------------------------------------------------------
Message: 1
Date: Sun, 9 Aug 2020 01:06:33 +0900
From: Austin Zhu <austinzhu...@gmail.com>
To: The Haskell-Beginners Mailing List - Discussion of primarily
beginner-level topics related to Haskell <beginners@haskell.org>
Subject: Re: [Haskell-beginners] Understanding functions like f a b c
= c $ b a
Message-ID:
<CAM2mjt-s5NxU0qcVP2oUDo4RYF_+R58C-7uya=y+p2xsiz+...@mail.gmail.com>
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Thank you for replying! It becomes a lot clearer now. :-)
On Sat, Aug 8, 2020, 14:25 Bob Ippolito <b...@redivi.com> wrote:
> I think the part that is confusing is that there are two steps here, there
> is the *foldr*, and then there is the application of *id* to the result
> of the *foldr*. *foldr* is of type *(a -> b -> b) -> b -> [a] -> b*, and
> in your example the type for *a* is *Integer* (probably not precisely
> Integer, but let's just say it is for simplicity) and the type for *b* is
> *[Integer]
> -> [Integer]*. It would be better to think of it as *(foldr f (const [])
> xs) id*. Another way to think of it is that *foldr* replaces the list *:*
> constructor with the function (*f*) and the *[]* constructor with the
> given *b* (*id*). Here's how I would think about the computation. In
> Haskell it's usually best to start with the outside and work in, due to the
> non-strict evaluation. At the end I've removed the bold from the terms that
> are already completely reduced.
>
> *init' [1, 2, 3]*
> *(foldr f (const []) (1 : 2 : 3 : [])) id*
> *(1 `f` (2 `f` (3 `f` const []))) id*
> *id ((2 `f` (3 `f` const [])) (1:))*
>
> *(2 `f` (3 `f` const [])) (1:)*
> 1 :* ((3 `f` const []) (2:))*
> 1 : 2 :* (const [] (3:))*
> 1 : 2 : []
>
>
> On Fri, Aug 7, 2020 at 7:12 AM Austin Zhu <austinzhu...@gmail.com> wrote:
>
>> Hello!
>>
>> I'm learning Haskell and I found an interesting implementation of init
>> using foldr. However I have difficulty understand how it works.
>>
>> *init' xs = foldr f (const []) xs id*
>> * where f x g h = h $ g (x:)*
>>
>> Consider I have a input of *[1,2,3]*, then is would become
>>
>> *f 1 (f 2 ( f 3 (const []))) id*
>>
>> I substitute those parameters into f and the innermost one becomes *h $
>> (const []) (1:)*, which is simply *h []*. However when I want to reduce
>> the expression further, I found it's hard to grasp. The next one becomes *f
>> 2 (h [])* , which is
>>
>> *h $ (h []) (2:)*
>>
>> if it works like that. This looks confusing to me. To match the type of
>> *foldr*, h should be of type *[a] -> [a]* and *h []* would just be of
>> type *[a]*, which isn't applicable to *(2:)*.
>>
>> I also thought it in another way that *f x g* returns a function of type
>> *([a]
>> -> [a]) -> [a],* this kinda makes sense considering applying *id*
>> afterwards. But then I realized I still don't know what this *h* is
>> doing here. It looks like *h* conveys *g (x:)* from last time into the
>> next application.
>> Did I miss something when I think about doing fold with function as
>> accumulator?
>>
>> I'd really appreciate if anyone could help me with this.
>> _______________________________________________
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>>
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