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

[michael rice]

I forgot the type signiture on the last last:

(~>>) :: (Seed -> (a,Seed)) -> (Seed -> (b,Seed)) -> (Seed -> (b,Seed))
(~>>) m n = \seed0 ->
  let (result1, seed1) = m seed0
      (result2, seed2) = n seed1
  in (result2, seed2)

========

With it I get this:

*Main> rollDie 362354 ~>> (rollDie ~>> rollDie)



<interactive>:1:0:

    Couldn't match expected type `Seed -> (a, Seed)'

           against inferred type `(Int, Seed)'

    In the first argument of `(~>>)', namely `rollDie 362354'

    In the expression: rollDie 362354 ~>> (rollDie ~>> rollDie)

    In the definition of `it':

        it = rollDie 362354 ~>> (rollDie ~>> rollDie)

*Main> 


Michael

--- On Wed, 4/22/09, Achim Schneider <barsoap at web.de> wrote:

From: Achim Schneider <barsoap at web.de>
Subject: [Haskell-cafe] Re: Overriding a Prelude function?
To: haskell-cafe at haskell.org
Date: Wednesday, April 22, 2009, 9:52 PM

michael rice <nowgate at yahoo.com> wrote:

> 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

Well, you obviously need an initial seed:

rollDie 0xdeadbeef ~>> (rollDie ~>> rollDie)

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