The code you gave looks fine to me, and indeed compiled. But to fill out the instance declaration you'll need to a) make Data the context in the App constructor b) make Data the context in the instance Data (Term a) declaration Also there is absolutely no point in the (Typeable a) context for the data declaration, so I dropped it. Here's a filled-out version that works: Simon module Foo where import Data.Typeable import Data.Generics data Term a = Const a | LVar Int | forall b. Data b => App (Term (b -> a)) (Term b) | Lam (Term a) instance (Typeable a) => Typeable (Term a) where typeOf w = mkAppTy (mkTyCon "Term.Term") [typeOf (undefined :: a)] instance (Data a) => Data (Term a) where toConstr (Const _) = mkConstr 1 "Const" Prefix toConstr (LVar _) = mkConstr 3 "LVar" Prefix toConstr (App _ _) = mkConstr 4 "App" Prefix toConstr (Lam _) = mkConstr 5 "Lam" Prefix gmapT f (Const a) = Const (f a) gmapT f (LVar i) = LVar (f i) gmapT f (App t1 t2) = App (f t1) (f t2) gmapT f (Lam t) = Lam (f t) | -----Original Message----- | From: glasgow-haskell-users-bounces@haskell.org [mailto:glasgow-haskell-users- | bounces@haskell.org] On Behalf Of Akos Korosmezey | Sent: 21 April 2004 13:23 | To: glasgow-haskell-users@haskell.org | Subject: Typeable and 'forall' in data constructors | | I am tying to write a Term class with function application: | | data (Typeable a) => Term a = | Const a | | LVar Int | | forall b. Typeable b => App (Term (b -> a)) (Term b) | | Lam (Term a) | | Because 'forall' is present, ghc refuses to derive Typeable and Data for | Term. I tried to implement them: | | instance (Typeable a) => Typeable (Term a) where | typeOf w = mkAppTy (mkTyCon "Term.Term") [typeOf (undefined :: a)] | | instance (Typeable a) => Data (Term a) where | toConstr (Const _) = mkConstr 1 "Const" Prefix | toConstr (LVar _) = mkConstr 3 "LVar" Prefix | toConstr (App _ _) = mkConstr 4 "App" Prefix | toConstr (Lam _) = mkConstr 5 "Lam" Prefix | | But ghc 6.2.1 returns with error on the line 'toConstr (App _ _)...': | parse error on input `b'. How can this be fixed? | Thank you | | Akos Korosmezey | | | | _______________________________________________ | Glasgow-haskell-users mailing list | Glasgow-haskell-users@haskell.org | http://www.haskell.org/mailman/listinfo/glasgow-haskell-users