Hi Chris, The change to ($)'s type is indeed intentional. The short answer is that ($)'s type prior to GHC 8.0 was lying a little bit. If you defined something like this: unwrapInt :: Int -> Int# unwrapInt (I# i) = i You could write an expression like (unwrapInt $ 42), and it would typecheck. But that technically shouldn't be happening, since ($) :: (a -> b) -> a -> b, and we all know that polymorphic types have to live in kind *. But if you look at unwrapInt :: Int -> Int#, the type Int# certainly doesn't live in *. So why is this happening? The long answer is that prior to GHC 8.0, in the type signature ($) :: (a -> b) -> a -> b, b actually wasn't in kind *, but rather OpenKind. OpenKind is an awful hack that allows both lifted (kind *) and unlifted (kind #) types to inhabit it, which is why (unwrapInt $ 42) typechecks. To get rid of the hackiness of OpenKind, Richard Eisenberg extended the type system with levity polymorphism [1] to indicate in the type signature where these kind of scenarios are happening. So in the "new" type signature for ($): ($) :: forall (w :: Levity) a (b :: TYPE w). (a -> b) -> a -> b The type b can either live in kind * (which is now a synonym for TYPE 'Lifted) or kind # (which is a synonym for TYPE 'Unlifted), which is indicated by the fact that TYPE w is polymorphic in its levity type w. Truth be told, there aren't that many Haskell functions that actually levity polymorphic, since normally having an argument type that could live in either * or # would wreak havoc with the RTS (otherwise, how would it know if it's dealing with a pointer or a value on the stack?). But as it turns out, it's perfectly okay to have a levity polymorphic type in a non-argument position [2]. Indeed, in the few levity polymorphic functions that I can think of: ($) :: forall (w :: Levity) a (b :: TYPE w). (a -> b) -> a -> b error :: forall (v :: Levity) (a :: TYPE v). HasCallStack => [Char] -> a undefined :: forall (v :: Levity) (a :: TYPE v). HasCallStack => a The levity polymorphic type never appears directly to the left of an arrow. The downside of all this is, of course, that the type signature of ($) might look a lot scarier to beginners. I'm not sure how you'd want to deal with this, but for 99% of most use cases, it's okay to lie and state that ($) :: (a -> b) -> a -> b. You might have to include a disclaimer that if they type :t ($) into GHCi, they should be prepared for some extra information! Ryan S. ----- [1] https://ghc.haskell.org/trac/ghc/wiki/NoSubKinds [2] https://ghc.haskell.org/trac/ghc/ticket/11473