Hi,
The types of `foldr` and `scanr` are different:

> :t scanr

scanr :: (a -> b -> b) -> b -> [a] -> [b]

> :t foldr

foldr :: (a -> b -> b) -> b -> [a] -> b

I believe the reason is that `scanr` does something like a prefix sum, https://en.wikipedia.org/wiki/Prefix_sum

So to get this to work, you need to change the type signature of filter''

filter'' :: (a -> Bool) -> [a] -> [[a]]

That's all.

Anyway you can try entering the definition of filter'' into ghci, and use :t to let ghc figure our the type for you as well :)

Best,

Zhi An

On Thu, May 1, 2014 at 10:42 AM, raffa f <freitasraffa@gmail.com> wrote:

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!

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