You can not convert an IO Int to Int, or at least, you shouldn't. However, you can do as follows:
test :: IO () test = do int <- randomRIO -- or whatever it is called print $ useInt int useInt :: Int -> Int useInt x = x+10 //Tobias 2009/6/9 ptrash <ptr...@web.de>: > > Hi, > > I am using the System.Random method randomRIO. How can I convert its output > to an Int? > > Thanks... > -- > View this message in context: > http://www.nabble.com/Convert-IO-Int-to-Int-tp23940249p23940249.html > Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com. > > _______________________________________________ > Haskell-Cafe mailing list > Haskell-Cafe@haskell.org > http://www.haskell.org/mailman/listinfo/haskell-cafe > -- Tobias Olausson tob...@gmail.com _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
