On 2009/06/09, at 19:33, Tobias Olausson wrote:
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
Or, you can lift pure function into IO. the below test' function
almost same as above test function. (But I used randomIO instead of
randomRIO because it seemed to be a typo :-)
test' = print =<< fmap useInt randomIO
I think it is more handy than using do notation, when you want to do
something simple with monads. And converting IO Int to IO anything is
much easier and safer than converting IO Int to Int.
ghci> :m +System.Random Data.Char
ghci> :t fmap (+1) randomIO
fmap (+1) randomIO :: (Num a, Random a) => IO a
ghci> :t fmap show randomIO
fmap show randomIO :: IO String
ghci> :t fmap chr randomIO
fmap Data.Char.chr randomIO :: IO Char
ghci> :t fmap (+) randomIO
fmap (+) randomIO :: (Num a, Random a) => IO (a -> a)
Thanks,
Hashimoto
//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...
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Tobias Olausson
tob...@gmail.com
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