#14010: UndecidableSuperClasses - Could not deduce (Category d) -------------------------------------+------------------------------------- Reporter: int-index | Owner: (none) Type: bug | Status: new Priority: normal | Milestone: Component: Compiler | Version: 8.0.1 Resolution: | Keywords: | UndecidableSuperClasses Operating System: Unknown/Multiple | Architecture: Type of failure: GHC rejects | Unknown/Multiple valid program | Test Case: Blocked By: | Blocking: Related Tickets: | Differential Rev(s): Wiki Page: | -------------------------------------+------------------------------------- Comment (by simonpj): GHC's error message looks absolutely right to me. Writing in the kinds we have {{{ Functor :: forall c d. (c -> d) -> Constraint OnSnd :: forall c d. ((c,d) -> *) -> c -> d -> * class (Category c, Category d) => MyFunctor @c @d (f :: c -> d) where instance MyFunctor @(c,d) @* (f2 :: (c, d) -> *) => MyFunctor @d @* (OnSnd @c @d f2 (a::c)) where }}} From the instance declaration we are obliged to generate superclasses for `MyFunctor @d @* (OnSnd @c @d f2 (a::c))`, namely {{{ [W] Category d [W] Category * }}} The latter is no problem, but `Category d` is. What do we have available to prove it? Just the instance context {{{ [G] MyFunctor @(c,d) @* (f2 :: (c,d) -> *) }}} If we have that given, then we have its superclasses given: {{{ [G] Category (c,d) [G] Category * }}} Can we prove `Category d` given `Category (c,d)`? No we cannot. And that is just what GHC says. Here it is with `-fprint-explicit-kinds`: {{{ T14010.hs:45:10: error: * Could not deduce (Category d) arising from the superclasses of an instance declaration from the context: MyFunctor (c, d) * f2 bound by the instance declaration at T14010.hs:(45,10)-(46,31) }}} So I claim this is a non-bug. -- Ticket URL: http://ghc.haskell.org/trac/ghc/ticket/14010#comment:2 GHC http://www.haskell.org/ghc/ The Glasgow Haskell Compiler