Hello,
I am just learning Haskell. Now, I encountered something that I cannot solve by myself. Your advice will be greatly appreciated.
Given a list of numbers, I want to modify each of those numbers by adding a random offset. However, each such modified number shall stay within certain bounds, given by the integers minValue and maxValue. After that, I want to continue computation with the resulting list of type [Int]. But for demonstration, I made a program that just prints out the list:
This program will not compile. GHC complains:import IO; import Random minValue = 0::Int maxValue = 1000::Int normalize a | a < minValue = minValue | a > maxValue = maxValue | otherwise = a modify a = do offset <- randomRIO(-100::Int, 100) return(normalize(a + offset)) main = putStrLn $ show $ map (modify) [0, 200, 400, 600, 800, 1000]
I understand that the result of the modify function is not an Int, as I would like to have it, but instead IO Int, and that cannot be applied to show. (I also did not quite understand why I need those brackets around the return value of the modify value. It won't compile if I leave them out, but I can accept that for now.)test.hs:14:18: No instance for (Show (IO Int)) arising from a use of `show' at test.hs:14:18-21 Possible fix: add an instance declaration for (Show (IO Int)) In the first argument of `($)', namely `show' In the second argument of `($)', namely `show $ map (modify) [0, 200, 400, 600, ....]' In the expression: putStrLn $ show $ map (modify) [0, 200, 400, 600, ....]
I also figured out how to generate a modified list of type [IO Int] and of type IO [Int]. However, I could not find out how to completely get rid of the IO monad and just get a mofied list of type [Int], which is what I really want.
Please, do You have any advice for me? I tried for some hours, and now I am really angry at that IO monad that sticks to my pretty integers like glue!
Also, any comment on the programming style and how I could achive my goals easier would be appreciated. (I left out comments and function types for the sake of brevity.)
Thanks a lot in advance.
Madoc.