Well, you can resort to functional dependencies, I guess... {-# LANGUAGE FlexibleInstances, FunctionalDependencies, MultiParamTypeClasses, UndecidableInstances #-} module FunDeps where data Rec a r = Rec a r data RecNil = RecNil data Wrapper a = Wrapper a class Wrapped r w | r -> w where i :: r -> w instance Wrapped RecNil RecNil where i RecNil = RecNil instance Wrapped r w => Wrapped (Rec a r) (Rec (Wrapper a) w) where 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) ...but that would be just an awkward way to do the same thing, so, my advice: don't. Type families are much nicer. On the other hand, you could do your "Rec" type polimorphic in "wrapper"; assuming your real-life Wrapper is not just an identity, this would be worth considering: {-# LANGUAGE KindSignatures #-} module PolyM where data Rec a r w = Rec (w a) (r w) data RecNil (w :: * -> *) = RecNil data Wrapper a = Wrapper a -- in reality it should be something else newtype Identity a = Identity a class Wrapped r where i :: r Identity -> r Wrapper instance Wrapped RecNil where i RecNil = RecNil instance Wrapped r => Wrapped (Rec a r) where i (Rec (Identity a) r) = Rec (Wrapper a) (i r) type TTest = Rec Int (Rec String RecNil) Identity type TTestWrapped = Rec Int (Rec String RecNil) Wrapper a :: TTest a = Rec (Identity 1) (Rec (Identity "a") RecNil) f :: TTestWrapped -> (Int, String) f (Rec (Wrapper n) (Rec (Wrapper s) RecNil)) = (n, s) r = f (i a) 24.11.2010 12:24, kg пишет:
Ok, it's exactly what i hoped.
And I would like to know (for fun) if it's possible to do it without type family extension. I've tried ... without success.
Thx.
On 11/22/2010 10:46 PM, Miguel Mitrofanov wrote:
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.
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