Re: [Haskell-cafe] Overriding a Prelude function?

22 Apr 2009

      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 ~]$
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