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Today's Topics:
1. Understanding functions like f a b c = c $ b a (Austin Zhu)
2. Re: Understanding functions like f a b c = c $ b a (Bob Ippolito)
3. Re: Understanding functions like f a b c = c $ b a
(Francesco Ariis)
4. Re: Understanding functions like f a b c = c $ b a
(Francesco Ariis)
----------------------------------------------------------------------
Message: 1
Date: Fri, 7 Aug 2020 23:11:43 +0900
From: Austin Zhu <austinzhu...@gmail.com>
To: beginners@haskell.org
Subject: [Haskell-beginners] Understanding functions like f a b c = c
$ b a
Message-ID:
<CAM2mjt_y2=rt_gz3vwh+e1xb+a1us+koakvhs1tskrpm3bx...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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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Message: 2
Date: Fri, 7 Aug 2020 22:24:57 -0700
From: Bob Ippolito <b...@redivi.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:
<cacwmpm98kn_p4gimdmpgwpy-jpqshyneu8tgvtk6989vxch...@mail.gmail.com>
Content-Type: text/plain; charset="utf-8"
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.
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
>
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Message: 3
Date: Sat, 8 Aug 2020 13:24:37 +0200
From: Francesco Ariis <fa...@ariis.it>
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: <20200808112437.GA15774@extensa>
Content-Type: text/plain; charset=us-ascii
+1
Il 07 agosto 2020 alle 22:24 Bob Ippolito ha scritto:
> 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.
> > _______________________________________________
> > Beginners mailing list
> > Beginners@haskell.org
> > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
> >
> _______________________________________________
> Beginners mailing list
> Beginners@haskell.org
> http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
------------------------------
Message: 4
Date: Sat, 8 Aug 2020 13:37:42 +0200
From: Francesco Ariis <fa...@ariis.it>
To: beginners@haskell.org
Subject: Re: [Haskell-beginners] Understanding functions like f a b c
= c $ b a
Message-ID: <20200808113742.GA22015@extensa>
Content-Type: text/plain; charset=utf-8
Il 08 agosto 2020 alle 13:24 Francesco Ariis ha scritto:
> +1
This is silly me thinking I was replying to a Discourse thread liking
the post…
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