How Lambda function works in Haskell

Question

I am new to Haskell, while working on small programs I found little confusions about functioning of lambda functions.

lastThat :: (a -> Bool) -> a -> [a] -> a   
lastThat f = foldl (\x acc -> if f x then x else acc)

Executing lastThat (>0) 100 [-1,-4,5,7,9,-10] I got 100 . While using the following definition

lastThat :: (a -> Bool) -> a -> [a] -> a   
lastThat f = foldl (\acc x -> if f x then x else acc)

& then executing lastThat (>0) 100 [-1,-4,5,7,9,-10] I got 9 as expected.

Why it did not worked with first definition?

Because in the former `x` is the *accumulator*, so you call `f` on the *accumulator*. — Willem Van Onsem, Dec 14 '19 at 12:57
when I use foldr with first definition as follows `firstThat :: (a-> Bool) -> a -> [a] -> a firstThat f = foldr (\x acc -> if f x then x else acc)` I got correct result as 5 but not 100. — Ayush kabir verma, Dec 14 '19 at 13:02
but the order of the accumulator and the element are swapped between `foldr` and `foldl`. — Willem Van Onsem, Dec 14 '19 at 13:02

score 0 · Answer 1 · answered Dec 22 '19 at 03:15

Both foldl & foldr traverses a list from left to right but the difference is foldl is Left associative and foldr is right associative

Given as list (xs)

xs = 1, 2, 3, ....., n

Think like foldl places the initial value to the left and associates the brackets to the left.

acc = 0
op = +
(((...(0 + 1) + 2) + 3) + 4) ..... + n)

foldr places the initial value to the right and associates the brackets to the right.

acc = 0
op = +
(1 + (2 + (3 + ......+ (n + 0)))...)))

1 Answers1