Haskell foldl and foldr implementation

fold_implementation.hs

foldl' :: (a -> b -> a) -> a -> [b] -> a
foldl' fn acc [] = acc
foldl' fn acc (x:xs) = foldl' fn (fn acc x) xs

foldr' :: (a -> b -> b) -> b -> [a] -> b
foldr' fn acc [] = acc
foldr' fn acc (x:xs) = fn x (foldr' fn acc xs)

f :: Double -> Double -> Double
f = \acc x -> acc / x + 1

main :: IO ()
main = do
  print (foldl' f  0 [1,2,3])
  print (foldr' f  0 [1,2,3])
  print (foldr' (/) 1 [2,4,8])
  

Your implementation of foldl' is incorrect because the function operation is not strict as required by the specification.
https://hackage.haskell.org/package/base-4.17.0.0/docs/Data-List.html#v:foldl-39-

You can test it using evaluate (foldl' (const id) () [throw StrictException, ()]) which is supposed to throw exception but won't. To fix, define as foldl' f z (x : xs) = let z' = f z x in z' seq foldl' f z' xs. See https://wiki.haskell.org/Seq

@asarkar the foldl' from Data.List you are referring to is indeed the strict version of foldl but the original creator of this post was clearly referring to the standard versions of foldl and foldr as he also said in the title of the post.