Re: [Haskell-cafe] Bug with QuickCheck 1.1 and GHC 6.8.2

On Thu, Aug 14, 2008 at 12:15 PM, Patrick Perry wrote:

Actually, a much better solution is:

variant :: Int -> Gen a -> Gen a variant v (Gen m) = Gen (\n r -> m n (rands r v)) where rands r0 0 = r0 rands r0 n = let (r1,r2) = split r0 (n',s) = n `quotRem` 2 in case s of 0 -> rands r1 n' _ -> rands r2 n'

It's not really clear to me how this works. Could you please explain? Cheers, Johan