Using the same basic structure you did, and foldr, I think below is the simplest method: ==================== import Data.Maybe searchList :: (a -> Bool) -> [a] -> Maybe [a] searchList p xs = foldr (\x acc -> if p x then Just (x: fromMaybe [] acc) else acc) Nothing xs ==================== ghci> searchList (=='o') "A quick brown fox" Just "oo" ghci> searchList (==' ') "A quick brown fox" Just " " ghci> searchList (=='z') "A quick brown fox" Nothing
From maybe gets rid of the Maybe, so that our recursive call works: ghci> fromMaybe [] (Just [1..3]) [1,2,3]
That's why we got the error below when we tried without fromMaybe; on subsequent applications of foldr, the type would have to change. ==================== <interactive>:1:51: Couldn't match expected type `[a]' against inferred type `Maybe [a]' In the expression: if p x then Just (x : acc) else acc In the first argument of `foldr', namely `(\ x acc -> if p x then Just (x : acc) else acc)' In the expression: foldr (\ x acc -> if p x then Just (x : acc) else acc) Nothing xs ==================== I have a feeling that using fromMaybe is not the best way, but it gets the job done for now. On that note; if somebody with some more experience would chime in, that'd be awesome. ;) Ezra dima.neg wrote:
How can I do this using foldr?
searchList p [] = Nothing searchList p (x:xs) | p x = Just (x:filter p xs) | otherwise = searchList p xs
I try this: searchList p xs = foldr (\x acc -> if p x then Just (x:acc) else acc) Nothing xs but don't work.
Thanks
-- View this message in context: http://old.nabble.com/Recursive-to-foldr-tp26368900p26399795.html Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.