hi everyone! here's my new problem. i wrote my version of filter: filter' :: (a -> Bool) -> [a] -> [a] filter' f = foldr (\x acc -> if f x then x:acc else acc) [] and it works! however, i wanted to use scan too. so i just replaced foldr with scanr, to see what would happen: filter'' :: (a -> Bool) -> [a] -> [a] filter'' f = scanr (\x acc -> if f x then x:acc else acc) [] but that doesn't work! ghci gives me this: folds.hs:15:59: Couldn't match expected type `a' with actual type `[a0]' `a' is a rigid type variable bound by the type signature for filter'' :: (a -> Bool) -> [a] -> [a] at folds.hs:14:13 In the second argument of `scanr', namely `[]' In the expression: scanr (\ x acc -> if f x then x : acc else acc) [] In an equation for filter'': filter'' f = scanr (\ x acc -> if f x then x : acc else acc) [] Failed, modules loaded: none. the problem seems to be with the start value of [], it seems? i don't understand, i thought scan and fold worked pretty much the same. i learned about these functions today, so i'm still trying to wrap my head around them... thank you!