Why can you reverse list with foldl, but not with foldr in Haskell

Question

Why can you reverse a list with the foldl?

reverse\' :: [a] -> [a]
reverse\' xs = foldl (\\acc x-> x : acc) [] xs

But this one gives me a

Answer 1

This is what foldl op acc does with a list with, say, 6 elements:

(((((acc `op` x1) `op` x2) `op` x3) `op` x4) `op` x5 ) `op` x6

while foldr op acc does this:

x1 `op` (x2 `op` (x3 `op` (x4 `op` (x5 `op` (x6 `op` acc)))))

When you look at this, it becomes clear that if you want foldl to reverse the list, op should be a "stick the right operand to the beginning of the left operand" operator. Which is just (:) with arguments reversed, i.e.

reverse' = foldl (flip (:)) []

(this is the same as your version but using built-in functions).

When you want foldr to reverse the list, you need a "stick the left operand to the end of the right operand" operator. I don't know of a built-in function that does that; if you want you can write it as flip (++) . return.

reverse'' = foldr (flip (++) . return) []

or if you prefer to write it yourself

reverse'' = foldr (\x acc -> acc ++ [x]) []

This would be slow though.