[Haskell-beginners] Understanding functions like f a b c = c $ b a

[Francesco Ariis]

+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 <austinzhu666 at 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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