michael rice wrote:

OK, I changed the operator from (>>) to (~>>). When I try to use it I get this:

[michael@localhost ~]$ ghci rand GHCi, version 6.10.1: http://www.haskell.org/ghc/__ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. [1 of 1] Compiling Main________________________ ( rand.hs, interpreted ) Ok, modules loaded: Main. *Main> rollDie ~>> (rollDie ~>> rollDie)

<interactive>:1:0: ______ No instance for (Show (Seed -> (Int, Seed))) __________ arising from a use of `print' at <interactive>:1:0-32 ______ Possible fix: __________ add an instance declaration for (Show (Seed -> (Int, Seed))) ______ In a stmt of a 'do' expression: print it

Well, you obviously need an initial seed: rollDie 0xdeadbeef ~>> (rollDie ~>> rollDie) -- (c) this sig last receiving data processing entity. Inspect headers for copyright history. All rights reserved. Copying, hiring, renting, performance and/or quoting of this signature prohibited.

*Main> :t rollDie ~>> (rollDie ~>> rollDie) rollDie ~>> (rollDie ~>> rollDie) :: Seed -> (Int, Seed) This is a function. How exactly do you want ghci to show it? When you figure that out, feel free to make an instance of Show for it. Meanwhile, you can just apply the function to a Seed value and see what happens: *Main> rollDie ~>> rollDie ~>> rollDie $ 6 (1,1145965850) michael rice wrote:

OK, I changed the operator from (>>) to (~>>). When I try to use it I get this:

[michael@localhost ~]$ ghci rand GHCi, version 6.10.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. [1 of 1] Compiling Main ( rand.hs, interpreted ) Ok, modules loaded: Main. *Main> rollDie ~>> (rollDie ~>> rollDie)

<interactive>:1:0: No instance for (Show (Seed -> (Int, Seed))) arising from a use of `print' at <interactive>:1:0-32 Possible fix: add an instance declaration for (Show (Seed -> (Int, Seed))) In a stmt of a 'do' expression: print it *Main>

Michael

--- On *Wed, 4/22/09, Luke Palmer //* wrote:

From: Luke Palmer Subject: Re: [Haskell-cafe] Overriding a Prelude function? To: "michael rice" Cc: "Ross Mellgren" , "Dan Weston" , "haskell-cafe@haskell.org" Date: Wednesday, April 22, 2009, 5:02 PM

On Wed, Apr 22, 2009 at 1:47 PM, michael rice > wrote:

Here's what I get:

[michael@localhost ~]$ ghci GHCi, version 6.10.1: http://www.haskell.org/ghc/ :? for help Loading package ghc-prim ... linking ... done. Loading package integer ... linking ... done. Loading package base ... linking ... done. Prelude> import Prelude hiding ((>>))

You know, to avoid this nonsense you could just name the operator something else, like >>~, or ~>>, or $@**!. Operators are just names.

Luke

<interactive>:1:0: parse error on input `import' Prelude>

=====

I was passing seed0 to rollDie and getting back (r1,seed1) passing seed1 to rollDie and getting back (r2,seed2) passing seed2 to rollDie and getting back (r3,seed3)

Just based on the problem text, I would guess that passing rollDie and seed0 to (>>) I would get back (r3,seed3), losing the intermediate random numbers r1 and r2 along the way, at least that's what I understood it to say.

So, I know that next I'm probably going to have to do something to remedy that, but I haven't gotten to that next step yet. What is unsugar?

Thanks in advance for your patience.

Michael

--- On *Wed, 4/22/09, Dan Weston />/* wrote:

From: Dan Weston > Subject: Re: [Haskell-cafe] Overriding a Prelude function? To: "Ross Mellgren" > Cc: "michael rice" >, "haskell-cafe@haskell.org " > Date: Wednesday, April 22, 2009, 12:37 PM

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 ~]$ >> >> >> _______________________________________________ >> Haskell-Cafe mailing list >> Haskell-Cafe@haskell.org http://mc/compose?to=Haskell-Cafe@haskell.org http://mc/compose?to=Haskell-Cafe@haskell.org>

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