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!