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 wrote: From: Achim Schneider Subject: [Haskell-cafe] Re: Overriding a Prelude function? To: haskell-cafe@haskell.org Date: Wednesday, April 22, 2009, 9:52 PM 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. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe