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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