#9123: Need for higher kinded roles -------------------------------------+------------------------------------ Reporter: simonpj | Owner: Type: bug | Status: new Priority: normal | Milestone: Component: Compiler | Version: 7.8.2 Resolution: | Keywords: Operating System: Unknown/Multiple | Architecture: Unknown/Multiple Type of failure: None/Unknown | Difficulty: Unknown Test Case: | Blocked By: Blocking: | Related Tickets: -------------------------------------+------------------------------------ Comment (by ekmett): {{{ newtype StateT s m a = StateT { runStateT :: s -> m (a, s) } }}} This means that when you go to describe the instance for `Representational (StateT s m)` it needs to convert `s -> m (a, s)` into `s -> m (b, s)` given a coercion between a and b. But that means we're coercing `(,) a` into `(,) b` -- let's call those f and g respectively -- and we're expanding that with the coercion from s into s which is trivial. If we can't lift `(Representable f, Coercion f g, Coercion a b)` into `Coercion (f a) (g b)` but can only go `(Representable f, Coercion a b) => Coercion (f a) (f b)` then we get stuck, as we only have `(Representable ((,) a), Coercion ((,) a) (,) b, Coercion s s)`! For state you could actually get by with a _different_ weaker rule added to the one Richard offered: {{{ ext :: forall (f :: x -> y) (g :: x -> y) (a :: x). Coercion f g -> Coercion (f a) (g a) }}} which I fake with `unsafeCoerce` right now in https://github.com/ekmett/roles/blob/master/src/Data/Roles.hs But that isn't enough to handle lifting a coercion where what 'a' occurs on both sides of the nested, which is what you get for free monads, cofree comonads, etc. The simplest case that fails is probably: {{{ newtype Pair a = Pair (a, a) }}} -- Ticket URL: http://ghc.haskell.org/trac/ghc/ticket/9123#comment:13 GHC http://www.haskell.org/ghc/ The Glasgow Haskell Compiler