Neil Mitchell wrote:
2 to p2 - to show how this can be done in the general case. With the specific information you know about >1 vs >2 you can do better, but
I suspected that you actually wanted to do something "cleverer" with the list, for the sake of argument, I'm going to change >1 to p1 and this gets across the general point:
f lst = show (sumPairs (>1) (>2) lst)
sumPairs :: (Int -> Bool) -> (Int -> Bool) -> [Int] -> (Int, Int) sumPairs p1 p2 [] = (0, 0) sumPairs p1 p2 (x:xs) = (add p1 a, add p2 b) where (a,b) = sumPairs xs add pred value = if pred x then x+value else value
[Untested, something like this should work]
Nope. That won't work because you end up creating huge "add" thunks which cause end up causing a stack overflow (tested with GHC -O2). I think you are probably going to need strictness in order to skin this cat in Haskell. Here's an example that does work... import Data.List main = print $ para_filter_sum (> 1) (> 2) lst twos = 2: twos lst = take 10000000 $ [1,2,3,4,5] ++ twos -- f lst = show (filter (> 1) lst, filter (> 2) lst) para_filter_sum f g xs = foldl' (\(n,m) elem -> seq n $ seq m $ (n+if f elem then elem else 0, m+if g elem then elem else 0 ) ) (0,0) xs Greg Buchholz