On Sun, Jul 20, 2008 at 10:30 PM, Rafael Gustavo da Cunha Pereira
Pinto
-------------sample.in------------- 01 02 03 04 05 06 10 11 15 18 29 45 19 22 10 01 23 14 -----------------------------------------
So, you got something like "10 11 15 18 29 45". So, we may want first to separate them into ["10", "11", "15", "18", "29", "45"]. Having this list, we simply apply read, constrained to String -> Int, to all its members, so we obtain [10, 11, 15, 18, 29, 45] :: [Int]. Now, translating what I said to Haskell: convert :: String -> [Int] convert = map read . words read :: FilePath -> IO [[Int]] read fp = (map convert . lines) `fmap` readFile fp (I've taken some "shortcuts" in the code above because that's how I'd write it. If you have trouble to understand it on a first glance, try to read carefully the docs of each function and mentally execute an example.) HTH, -- Felipe.