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You can reach the person managing the list at beginners-ow...@haskell.org When replying, please edit your Subject line so it is more specific than "Re: Contents of Beginners digest..." 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> Content-Type: text/plain; charset="utf-8" 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. >> _______________________________________________ >> 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 > -------------- next part -------------- An HTML attachment was scrubbed... URL: <http://mail.haskell.org/pipermail/beginners/attachments/20200809/0b339c0e/attachment-0001.html> ------------------------------ Subject: Digest Footer _______________________________________________ Beginners mailing list Beginners@haskell.org http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners ------------------------------ End of Beginners Digest, Vol 146, Issue 2 *****************************************