Hi,
--multRepl :: String -> [String] -> [String] -> [String]
Personally I woud use a list of pairs to represent the target/replacements: multRepl :: String -> [(String,String)] -> String That way you are guaranteed that for each target there is a replacement and don't need to handle the cases where both lists have different lengths. multRepl str srs = foldl (\acc (s,r) -> replace acc s r) str srs or in point-free form: multRepl = foldl (\acc (s,r) -> replace acc s r) or more succinctly (as mentionned before): multRepl = foldl (uncurry . replace) Patrick
multRepl [] _ _ = [] multRepl str (s:ss) (r:rs) = do let newStr = U.replace str s r if (length ss) == 0 then return newStr else multRepl newStr ss rs =======
this does produce the correct output: This is very original structure
and here are my questions:
1. the type *Main Useful> :t multRepl multRepl :: (Eq t) => [t] -> [[t]] -> [[t]] -> [[t]]
but i have it returning newStr which equals U.replace str s r and the type of U.replace is String as shown below *Main Useful> :t Useful.replace Useful.replace :: (Eq a) => [a] -> [a] -> [a] -> [a]
so why is it returning [String] when newStr isn't a list of strings?
2. is the way i've done it proper haskellian? it took me quite some time to think this out trying to find my way through the fog of imperative programming.
(my apologies for replying to my own posts - as well as my appreciation for your assistance)
-- In friendship, prad
... with you on your journey Towards Freedom http://www.towardsfreedom.com (website) Information, Inspiration, Imagination - truly a site for soaring I's
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-- ===================== Patrick LeBoutillier Rosemère, Québec, Canada