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

Francesco Ariis fa-ml at ariis.it
Mon Aug 10 01:05:59 UTC 2020


+1

Il 10 agosto 2020 alle 08:00 Kim-Ee Yeoh ha scritto:
> On Fri, Aug 7, 2020 at 9:12 PM 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:)*
> >
> >
> The last line isn’t correct because of erroneous alpha capture.
> 
> Modulo certain things that aren’t relevant here, the definition of the
> folding function f is equivalent to the eta-expansion: f x g = \h -> h (g
> (x:)). Note the lambda abstraction.
> 
> Try substituting that in
> 
> *f 1 (f 2 ( f 3 (const []))) id*
> 
> to see what you get.
> 
> Hint: Note how f 3 (const []) evaluates to
> 
> \h -> h (const [] (3:))
> = \h -> h []
> = ($ [])
> 
> Next f 2 ($ []) becomes
> 
> \h -> h (($ []) (2:))
> = \h -> h (2:[])
> = ($ (2:[]))
> 
> And you can see how you end up with init’ [1,2,3] = 1:(2:[]) = [1,2].
> Notice how I converted from a lambda abstraction to combinator form to
> prevent the named lambda variable h from obscuring what’s really going on.
> 
> Another way to figure out this out is by calculating the precise type of
> the folding function f that is provided to foldr and hence the type to h.
> 
> 
> 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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> > Beginners at haskell.org
> > http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
> >
> -- 
> -- Kim-Ee

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