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:
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]
This program will not compile. GHC complains:
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 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.)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.