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
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