I don't think you can get a type equality comparison test into type families without additional compiler support. If you are willing to restrict your labels to type-level naturals or some other closed universe, and allow undecidable instances, you can do something like this: data Z = Z data S a = S a type family Select label record type instance Select lbl (rlbl, ty, rest) = IfEq lbl rlbl ty (Select lbl rest) type family IfEq n0 n1 t f type instance IfEq Z Z t f = t type instance IfEq Z (S n) t f = f type instance IfEq (S n) Z t f = f type instance IfEq (S n0) (S n1) t f = IfEq n0 n1 t f Better support for closed type families that allowed overlap would be quite useful. -- ryan 2008/12/11 Taru Karttunen :

Hello

What is the correct way to transform code that uses record selection with TypeEq (like HList) to associated types? I keep running into problems with overlapping type families which is not allowed unless they match.

The fundep code:

class Select rec label val | rec label -> val instance TypeEq label label True => Select (Label label val :+: rest) label val instance (Select tail field val) => Select (any :+: tail) field val

And a conversion attempt:

class SelectT rec label where type S rec label instance TypeEq label label True => SelectT (Label label val :+: rest) label where type S (Label label val :+: rest) label = val instance (SelectT tail field) => SelectT (any :+: tail) field where type S (any :+: tail) field = S tail field

which fails with:

Conflicting family instance declarations: type instance S (Label label val :+: rest) label -- Defined at t.hs:19:9 type instance S (any :+: tail) field -- Defined at t.hs:23:9

How is it possible to get the TypeEq constraint into the type family?

Attached is a complete example that illustrates the problem.

- Taru Karttunen

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