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
==============
[mich...@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)
[mich...@localhost ~]$
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