Re: [Haskell-cafe] Convert IO Int to Int

On Tue, Jun 9, 2009 at 16:14, Daniel Fischer wrote:

Am Dienstag 09 Juni 2009 15:57:24 schrieb Magnus Therning:

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On Tue, Jun 9, 2009 at 2:52 PM, ptrash wrote:

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Hmm...it am not getting through it. I just want to generate a random number and then compare it with other numbers. Something like

r = randomRIO (1, 10) if (r > 5) then... else ...

You have to do it inside the IO monad, something like

    myFunc  = do         r <- randomRIO (1, 10         if r > 5             then ...             else ...

/M

Or make the source of the pseudo-random numbers explicit:

import System.Random

function :: (RandomGen g, Random a) => g -> other args -> result function gen whatever    | r > 5     = blah newgen something    | r < 3     = blub newgen somethingElse    | otherwise = bling      where        (r,newgen) = randomR (lo,hi) gen

and finally, when the programme is run:

main = do    args <- getArgs    sg <- getStdGen    foo <- thisNThat    print $ function sg foo

If you're doing much with random generators, wrap it in a State monad.

To avoid reinventing the wheel one can use excellent package available on Hackage: http://hackage.haskell.org/cgi-bin/hackage-scripts/package/MonadRandom

The die function simulates the roll of a die, picking a number between 1 and 6, inclusive, and returning it in the Rand monad. Notice that this code will work with any source of random numbers g.

die :: (RandomGen g) => Rand g Int die = getRandomR (1,6)

The dice function uses replicate and sequence to simulate the roll of n dice.

dice :: (RandomGen g) => Int -> Rand g [Int] dice n = sequence (replicate n die)

To extract a value from the Rand monad, we can can use evalRandIO.

main = do values <- evalRandIO (dice 2) putStrLn (show values)

Best regards Krzysztof Skrzętnicki