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

[Kim-Ee Yeoh]

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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-- 
-- Kim-Ee
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