Hi, Let us consider the following example: ----------------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a instance FooClass Integer where foo1 v = 1 foo2 v = 2 data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show instance FooClass Bar where foo1 b = case b of Bar1 i -> Bar1 (foo1 i) Exp1 b1 b2 -> Exp1 (foo1 b1) (foo1 b2) Exp2 b1 b2 -> Exp2 (foo1 b1) (foo1 b2) foo2 b = case b of Bar1 i -> Bar1 (foo2 i) Exp1 b1 b2 -> Exp1 (foo2 b1) (foo2 b2) Exp2 b1 b2 -> Exp2 (foo2 b1) (foo2 b2) main = do let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b print c print $ foo1 c print $ foo2 c ----------------------- We obtain as expected: $ runghc propagate_with_duplicated_code.hs Exp1 (Exp2 (Bar1 3) (Bar1 4)) (Bar1 4) Exp1 (Exp2 (Bar1 1) (Bar1 1)) (Bar1 1) Exp1 (Exp2 (Bar1 2) (Bar1 2)) (Bar1 2) My question is related to the code inside the Fooclass instance definition for Bar: we have repeated code where only "foo1" or "foo2" changes. So the first idea is to write a higher-order function, but it does not work: ----------------------- class FooClass a where foo1 :: a -> a foo2 :: a -> a instance FooClass Integer where foo1 v = 1 foo2 v = 2 data Bar = Bar1 Integer | Exp1 Bar Bar | Exp2 Bar Bar deriving Show propagate :: FooClass a => a -> (a->a) -> a propagate v f = case v of Bar1 i -> Bar1 (f i) Exp1 b1 b2 -> Exp1 (f b1) (f b2) Exp2 b1 b2 -> Exp2 (f b1) (f b2) instance FooClass Bar where foo1 b = propagate b foo1 foo2 b = propagate b foo2 main = do let a = Bar1 3 let b = Bar1 4 let c = Exp1 (Exp2 a b) b print c print $ foo1 c print $ foo2 c ----------------------- The problem is that the type variable `a` in the definition of `propagate` cannot match both the type of i, i.e. an integer, and the type of b1 and b2, i.e. Bar. Putting the function propagate in the typeclass does not help. How to factorize my code? Thanks in advance, TP