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