[Haskell-cafe] Re: so how does one convert an IO a into an a ?

André Pang ozone at algorithm.com.au
Thu Jul 8 19:24:25 EDT 2004

On 09/07/2004, at 4:50 AM, Crypt Master wrote:

> 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*

Other people have covered a lot about IO, but for your particular  
problem of random numbers, here's a reasonably simple solution:

     module RandomList where

     import Random

     seed :: Int
     seed = 69

     randomList :: [Int]
     randomList = randomRs (1,6) (mkStdGen seed)


     RandomList> :t randomList
     randomList :: [Int]
     RandomList> take 10 randomList

The key to figuring out how on earth to use the combinations of  
randomRs and generators is having good documentation on the Random  
module, which I found here:


I'm guessing you're using hugs, which does give you the Random module,  
but it's not exactly easy to figure out from reading the source code  
(especially if you're a Haskell beginner)!

% Andre Pang : trust.in.love.to.save

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