On 08/19/2011 08:50 AM, Ryan Ingram wrote:
ki1 :: KI () Int ki1 = KI @Int (\() s -> (s+1, s)) ki2 :: KI () Int ki2 = KI @() (\() () -> ((), 0)) f :: Bool -> KI () Int f x = if x then ki1 else ki2 iso f = KI ?? ?? The problem is that we have multiple possible internal state types!
Aha! Nice counterexample. Thanks :) _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
