A testcase make myFoldl in homework4 fail

[akamayu-ouo]

Hi, I find a testcase that make the myFoldl function fail.

(foldr . flip) (++) "A" ["B","C","D"]    -- outputs "ADCB", rather than "ABCD"

What's wrong is after flipping the reduction function, the arguments are swapped, but the order of application is still from right to left.

More explicit,

(foldr . flip) f x0 [x1, x2, x3, x4, x5, ... , xn]
= (f ( ... (f (f x0 xn) xn-1) ... ) x1)

while

foldl f x x0 [x1, x2, x3, x4, x5, ... , xn]
= (f ( ... (f (f x0 x1) x2) ... ) xn)

[bschnitz]

Correct. If the operations commutate, then the myFoldl definition from the solution does work. Another noncommutative example would be (foldr . flip) (mod) 4 [3, 2] -- outputs 0 whilst 1 would be correct To mimic the correct behaviour, one has also to reverse the input list to have the function application in the correct order:

myFoldl :: (b -> a -> b) -> b -> [a] -> b
myFoldl f base xs = foldr (flip f) base (reverse xs)