| iterIO uses mkTyCon for the simple reason that ((Typeable t, Typeable | m) => Iter t m) is Typeable1 and there is no automatic way of deriving | Typeable1.
This email is triggered by a thread on Haskell Cafe about changes to the Typeable class http://www.mail-archive.com/haskell-cafe@haskell.org/msg91830.html It proposes a modification to the way Typeable is derived; and concludes with a question. Simon David Mazieres and others comment that you can't derive Typeable for types like this: data T f = MkT (f Int) So he defines his own instance like this [C] instance Typable1 f => Typeable (T f) where typeOf = ... So why can't GHC do this? Well, here's what GHC does. Given a bog standard data type like Maybe data Maybe a = Nothing | Just a deriving( Typeable ) GHC generates this instance [A] instance Typeable1 Maybe where typeOf = ... Remember that Typeable1 takes a type *constructor*, of kind (*->*), as its argument. Now if we need (Typeable (Maybe Int)), GHC first uses an instance from the Typeable library: [B] instance (Typeable1 f, Typeable a) => Typeable (f a) where typeOf = ... And now it uses the (Typeable1 Maybe) instance [A]. So it's kind of cool... the applications are decomposed by [B], leaving the tycon to [A]. But this doesn't work for T above. We can't make (Typeable1 T) because T has kind ((*->*)->*), not (*->*) as Typeable1 requires. Hence David defining his own instance. GHC could do this too. Indeed it could do so for Maybe too, thus: instance Typeable a => Typeable (Maybe a) where typeOf = ... But then, alas, we could not get (Typeable (T Maybe)), because [C] needs Maybe to be in Typeable1. ========== PROPOSAL ============== So here is a compromise, which would at least do better than the current story: When deriving Typeable for a data type S of kind S :: k1 -> .. -> kn -> * -> ... -> * (where kn is not *, and there are M trailing * arguments), generate the instance instance (Typeable_x1 a1, ..., Typeable_xn an) => TypeableM (S a1 .. an) That is, knock off all the trailing * args, and then generate an instance for the remaining stub. ===================== EXAMPLE ============ Example from iterIO: newtype Iter (t :: *) (m :: *->*) (a :: *) = Iter { runIter :: Chunk t -> IterR t m a } deriving( Typeable ) This should generate instance (Typeable t, Typeable1 m) => Typeable1 (Iter t m) where we knock off the trailing (a :: *) argument. ================== QUESTION ================= This approach is not beautiful. It does not solve the underlying problem, which is a lack of kind polymorphism, but that is a battle for another day. Until that day, this alternative way of deriving Typeable would automate significantly more cases, I think. Of course, it also makes it more complicated to explain when "deriving Typeable" will succeed. Any opinions? Does anyone care? Simon _______________________________________________ Glasgow-haskell-users mailing list Glasgow-haskell-users@haskell.org http://www.haskell.org/mailman/listinfo/glasgow-haskell-users
