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
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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