True enough -- if you really want to redefine the monadic operator, you have to use {-# LANGUAGE NoImplicitPrelude #-} import Prelude hiding ((>>), (>>=), return) or something like it, although Michael's example didn't appear to be going quite that far. -Ross On Apr 22, 2009, at 12:37 PM, Dan Weston wrote:

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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Hahah yeah of course, I left it implicit that you'd only do this if you were changing the types (e.g. parameterized monads or what have you) -Ross On Apr 23, 2009, at 5:15 AM, Heinrich Apfelmus wrote:

Ross Mellgren wrote:

...

True enough -- if you really want to redefine the monadic operator, you have to use

{-# LANGUAGE NoImplicitPrelude #-}

import Prelude hiding ((>>), (>>=), return)

or something like it, although Michael's example didn't appear to be going quite that far.

Or just make it an instance of Monad . :D

Regards, apfelmus

-- http://apfelmus.nfshost.com

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