Sure, it's possible with TypeFamilies. The following compiles OK: {-# LANGUAGE TypeFamilies #-} module TypeCalc where data Rec a r = Rec a r data RecNil = RecNil data Wrapper a = Wrapper a class TypeList t where type Wrapped t i :: t -> Wrapped t instance TypeList RecNil where type Wrapped RecNil = RecNil i RecNil = RecNil instance TypeList r => TypeList (Rec a r) where type Wrapped (Rec a r) = Rec (Wrapper a) (Wrapped r) i (Rec a r) = Rec (Wrapper a) (i r) type TTest = Rec Int (Rec String RecNil) type TTestWrapped = Rec (Wrapper Int) (Rec (Wrapper String) RecNil) a :: TTest a = Rec 1 (Rec "a" RecNil) f :: TTestWrapped -> (Int, String) f (Rec (Wrapper n) (Rec (Wrapper s) RecNil)) = (n, s) r = f (i a) -- so, "i a" is of the type TTestWrapped. On 22 Nov 2010, at 23:43, kg wrote:

Hi,

I've tried to simplify as much as possible my problem. Finally, I think I can resume it like that:

Suppose these following data types : data Rec a r = Rec a r data RecNil = RecNil data Wrapper a = Wrapper a

Then, we can build the following type: type TTest = Rec Int (Rec String RecNil) or this type: type TTestWrapped = Rec (Wrapper Int) (Rec (Wrapper String) RecNil)

Is it possible to build TTestWrapped from TTest ?

Thx in advance, Antoine.

_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe