[Haskell-cafe] Re: Generating repeatable arbitrary values with QuickCheck 2

[Sean Leather]

On Tue, Feb 2, 2010 at 20:25, David Menendez wrote:

> On Tue, Feb 2, 2010 at 1:48 PM, Ryan Ingram wrote:
> > Gen slightly breaks the monad laws:
> >
> >> arbitrary >>= return
> > is not the same as
> >> return () >>= const arbitrary
> > because each bind splits the generator, so you end up with a different
> > seed passed to arbitrary in these two cases.
>

Ah yes, that was exactly the problem.


>  > If the observable value is "some random object" this is a safe fudge,
> > but if you want repeatable, it doesn't quite work.  You need your
> > instances to be exactly identical, down to the associativity of binds,
> > in order to get the same results.
>
> We could avoid that problem by redefining Gen as a state transformer monad.
>
> newtype Gen a = MkGen { unGen :: StdGen -> Int -> (a, StdGen) }
>
> instance Monad Gen where
>    return a = MkGen $ \r _ -> (a,r)
>    MkGen m >>= k = MkGen $ \r n -> let (a,r') = m r n in unGen (k a) r' n
>
> I'm pretty sure all the Gen primitives can be similarly redefined.
>

And I'm guessing I haven't been or won't be the only one running into this
issue.

Sean
-------------- next part --------------
An HTML attachment was scrubbed...
URL: http://www.haskell.org/pipermail/haskell-cafe/attachments/20100202/87ce07c9/attachment.html