Hi Thanks for your help so far, but I am still not getting this IO stuff. After reading your previous help and reading several articles on it I still cant phathom how you convert the IO Int into an Int. One person mentioned how random just returns an interative program which when eveluated returns the Int. Also from the school of expression book he says " The right way to think of (>>=) above is simply this: It "Executes" e1 ..." in relation to "do pat <- e1 ...". so I have this: <code> rollDice :: IO Int rollDice = getStdRandom (randomR (1,6)) rl :: [Int] rl = [ (getRndNum x) | x <- [1..] ] getRndNum :: Int -> Int getRndNum x = do n <- rollDice return n </code> *PS Pretend return is correctly aligned under n. dont what ahppens in copy and paste* now my understanding therefore is that "do n <- rollDice" should execute the the itererative program returned by rollDice. So now n should be my Int since IO Int was a program which when evaluted returns an Int ? However this is what haskell thinks of my thoery: *** Term : getRndNum *** Type : Int -> IO (Maybe Int) *** Does not match : Int -> Int So I am still in IO Int land despite having used the >>= in the do syntax. Worse Still I am in IO (Maybe Int) land. Monads within Monads. In yours, and many other examples I found online, the results are always passed to print which seems to know how to deal with an IO Int. Is this specially coded or overloaded or something ? There are plenty of examples which use return like so: do k <- getKey w return k which is what I tried above to no avail. It seems awefully complicated just to get hold a simple Int, but naturally complicity is directly related to ones understanding. Mine is sumewhat lacking ... any help would be appreciated. Thanks, S _________________________________________________________________ Tired of spam? Get advanced junk mail protection with MSN 8. http://join.msn.com/?page=features/junkmail