[Haskell-cafe] Overriding a Prelude function?
Thomas van Noort
thomas at cs.ru.nl
Wed Apr 22 15:23:50 EDT 2009
You can hide (>>) from the implicit import of Prelude using:
import Prelude hiding ((>>))
Kind regards,
Thomas
michael rice wrote:
> I've been working through this example from:
> http://en.wikibooks.org/wiki/Haskell/Understanding_monads
>
> I understand what they're doing all the way up to the definition of
> (>>), which duplicates Prelude function (>>). To continue following the
> example, I need to know how to override the Prelude (>>) with the (>>)
> definition in my file rand.hs.
>
> Michael
>
> ==============
>
> [michael at localhost ~]$ cat rand.hs
> import System.Random
>
> type Seed = Int
>
> randomNext :: Seed -> Seed
> randomNext rand = if newRand > 0 then newRand else newRand + 2147483647
> where newRand = 16807 * lo - 2836 * hi
> (hi,lo) = rand `divMod` 127773
>
> toDieRoll :: Seed -> Int
> toDieRoll seed = (seed `mod` 6) + 1
>
> rollDie :: Seed -> (Int, Seed)
> rollDie seed = ((seed `mod` 6) + 1, randomNext seed)
>
> sumTwoDice :: Seed -> (Int, Seed)
> sumTwoDice seed0 =
> let (die1, seed1) = rollDie seed0
> (die2, seed2) = rollDie seed1
> in (die1 + die2, seed2)
>
> (>>) m n = \seed0 ->
> let (result1, seed1) = m seed0
> (result2, seed2) = n seed1
> in (result2, seed2)
>
> [michael at localhost ~]$
>
>
>
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>
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