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

You can use foldr to reverse a list efficiently (well, most of the time in GHC 7.9—it relies on some compiler optimizations), but it's a little weird:

reverse xs = foldr (\x k -> \acc -> k (x:acc)) id xs []

I wrote an explanation of how this works on the Haskell Wiki.