[Haskell-cafe] Overriding a Prelude function?
michael rice
nowgate at yahoo.com
Wed Apr 22 21:22:09 EDT 2009
OK, I changed the operator from (>>) to (~>>). When I try to use it I get this:
[michael at 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 <lrpalmer at gmail.com> wrote:
From: Luke Palmer <lrpalmer at gmail.com>
Subject: Re: [Haskell-cafe] Overriding a Prelude function?
To: "michael rice" <nowgate at yahoo.com>
Cc: "Ross Mellgren" <rmm-haskell at z.odi.ac>, "Dan Weston" <westondan at imageworks.com>, "haskell-cafe at haskell.org" <haskell-cafe at haskell.org>
Date: Wednesday, April 22, 2009, 5:02 PM
On Wed, Apr 22, 2009 at 1:47 PM, michael rice <nowgate at yahoo.com> wrote:
Here's what I get:
[michael at 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 <westondan at imageworks.com> wrote:
From: Dan Weston <westondan at imageworks.com>
Subject: Re: [Haskell-cafe] Overriding a Prelude function?
To: "Ross Mellgren" <rmm-haskell at z.odi.ac>
Cc: "michael rice" <nowgate at yahoo.com>, "haskell-cafe at haskell.org" <haskell-cafe at 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 at 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 at localhost ~]$
>>
>>
>> _______________________________________________
>> Haskell-Cafe mailing list
>> Haskell-Cafe at haskell.org <mailto:Haskell-Cafe at haskell.org>
>> http://www.haskell.org/mailman/listinfo/haskell-cafe
>
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