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

[André Pang]

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)

Usage:

     RandomList> :t randomList
     randomList :: [Int]
     RandomList> take 10 randomList
     [6,2,6,2,6,2,1,3,2,3]
     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:

      
http://www.haskell.org/ghc/docs/latest/html/libraries/base/ 
System.Random.html#t%3ARandom

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