How does the foldr function in haskell work in this case?

Question

So I'm new to haskell and I kind of encountered this following expression which i don't quiet get how it works: foldr (.) (+3) [(*2), (+5)] 13 it gives out the result: 42 now i know that foldr normally works in an example like: foldr (+) 0 [1,2,3] like: (1+(2+(0+3))) but with adding another function (.) I kind of got confused. So please if any of you could explain to me exactly how haskell interprets this expression that would be great, thanks!

Answer 1

Comments might have already solved this for you. However, if not:

(.) is function composition:

f . g
= \x -> f $ g $ x
= \x -> f (g x).

Forms like (* 2) are sugar for functions of the form \\x -> x * 2

Now, observe that

foldr op base [a,b]

is the same as

a `op` (b `op` base)

of course, this works for higher numbers of arguments as well. For instance,

foldr (+) 0 [1,2,3,4,5]

is just

1 + 2 + 3 + 4 + 5 + 0

In your case, you ask about

foldr (.) (+3) [(*2), (+5)]

which is (for Integer )

(*2) . (+5) . (+3)
= \x -> (*2) $ (+5) $ (+3) $ x
= \x -> (*2) $ (+5) $ x + 3
= \x -> (*2) $ x + 3 + 5
= \x -> (x + 3 + 5) * 2
= \x -> x*2 + 16

and so

foldr (.) (+3) [(*2), (+5)] 13
= (\x -> x*2 + 16) 13
= 13*2 + 16
= 26 + 16
= 42

Answer 2

foldr fz can be seen as replacing the "cons" (:) of the list with f , and the empty list with z . This thus means that foldr fz [x 1 , x 2 … x n ] is equivalent to fx 1 (fx 2 ( … (fx n z) … ))

[(*2), (+5)] is syntactical sugar for (:) (*2) ((:) (+5) []) , so we can replace this with: (.) (*2) ((.) (+5) (+3)) , which is a verbose form of (*2) . (+5) . (+3) (*2) . (+5) . (+3) (*2) . (+5) . (+3) . This is thus a function. If we make a function application with this function and 13 as argument, we get:

   ((*2) . (+5) . (+3)) 13
-> ((*2) . (+5)) 16
-> (*2) 21
-> 42

foldl fz [x 1 , x 2 … x n ] is equivalent to f (… (f (fzx 1 ) x 2 ) … ) x n . So here that means foldl (.) (+3) [(*2), (+5)] is equivalent to (.) ((.) (+3) (*2)) (+5) , or less verbose (+3) . (*2) . (+5) (+3) . (*2) . (+5) (+3) . (*2) . (+5) . If we evaluate this for 13 , we obtain:

   ((+3) . (*2) . (+5)) 13
-> ((+3) . (*2)) 18
-> (+3) 36
-> 39

Answer 3

In Haskell (and in lambda calculus) you can do Eta reduction

In the example you shown, the long way to write would be:

foldr (\f h -> f . h) (\x -> x+3) [(\x -> x*2), (\x->x+5)] 13

Now you can apply Eta reduction one by one like this:

(\f h -> f . h)
(\f -> (f.))
(.)

next:

(\x -> x+3)
(+3)

(\\x -> x*2), (\\x->x+5) are so similar to (+3)

so finally you get you equivalent expression:

foldr (.) (+3) [(*2), (+5)]

and that means, compose each function, and when the list is empty, apply (+3)