Re: [Haskell-cafe] Existential question

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 :)