Be aware that the do unsugars to (Prelude.>>), not your (>>), even if you hide (Prelude.>>): import Prelude hiding ((>>)) m >> f = error "Call me!" main = putStrLn . show $ do [3,4] [5] The desugaring of the do { [3,4]; [5] } is (Prelude.>>) [3,4] [5] = [5,5], whereas you might have hoped for [3,4] >> [5] = error "Call me!" Dan Ross Mellgren wrote:
I think
import Prelude hiding ((>>))
does that.
-Ross
On Apr 22, 2009, at 11:44 AM, 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@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@localhost ~]$
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