This is a consequence of the way GHC generics represents datatypes that compose functor-like types in this fashion. If you compile that code with -ddump-deriv, you'll see that the Rep1 for Compose is (in abbreviated form): type Rep1 (Compose f g) = ... (f :.: Rec1 g) where (:.:) is defined as [1]: newtype (:.:) (f :: k2 -> *) (g :: k1 -> k2) (p :: k1) = Comp1 (f (g p)) In other words, we must kind-check the type (f (Rec1 g p)). But Rec1 is a datatype, so it must have result kind *. Therefore, the kind of f is forced to be (* -> *). I describe this in a Note here [2]. This feels like a somewhat fundamental consequence of using datatypes to abstract over a datatype's structure, so I'm not aware of a way around this. Luckily, the kind of g is still (k -> *), which was the main goal of Trac #10604 [3]. Ryan S. ----- [1] http://git.haskell.org/ghc.git/blob/0676e68cf5fe8696f1f760fef0f35dba14db1104... [2] http://git.haskell.org/ghc.git/blob/0676e68cf5fe8696f1f760fef0f35dba14db1104... [3] https://ghc.haskell.org/trac/ghc/ticket/10604 On Wed, Jun 1, 2016 at 10:55 AM, Simon Peyton Jones wrote:

Ryan

If you compile

newtype Compose f g a = Compose (f (g a)) deriving( Generic1 )

and do –show-iface on the resulting hi file, you’ll see

$fGeneric1Compose ::

forall (f :: * -> *) k (g :: k -> *).

Functor f =>

Generic1 (Compose f g)

I was expecting to see

$fGeneric1Compose ::

forall (f :: k1 -> *) k (g :: k -> k1).

(..something..) =>

Generic1 (Compose f g)

Otherwise the Generic1 instance only works if its first argument has kind (* -> *).

Is that the intention? Maybe so…

Simon