Functional dependencies with Type Classes
Hi, suppose there are two (identical) classes:
class Res a b | a -> b where getRes :: a -> b
and
class Res2 t where type Member t getRes2 :: t -> Member t
It is easy to automatically make every instance of Res2 an instance of res:
instance Res2 a => Res a (Member a) where getRes x = getRes2 x
However, declaring every instance of Res an instance of Res2 seems impossible, as the following doesn't compile
instance Res a b => Res2 a where type Member a = b getRes2 x = getRes x
Question is: How to do this? The reason I need it is because I use a library which uses functional dependencies, but my classes shall be type families. Regards, Henning
Am Freitag, 28. März 2008 02:12 schrieb Henning Günther:
Hi,
suppose there are two (identical) classes:
class Res a b | a -> b where getRes :: a -> b
and
class Res2 t where type Member t getRes2 :: t -> Member t
It is easy to automatically make every instance of Res2 an instance of
res:
instance Res2 a => Res a (Member a) where getRes x = getRes2 x
However, declaring every instance of Res an instance of Res2 seems impossible, as the following doesn't compile
instance Res a b => Res2 a where type Member a = b getRes2 x = getRes x
Question is: How to do this? The reason I need it is because I use a library which uses functional dependencies, but my classes shall be type families.
Regards, Henning
Hello Henning, I also came across this problem half a year ago, tried to find a solution and came to the conclusion that there is none. It seems as if functional dependencies are a trap: Once you’ve started using them it is impossible to escape them by switching to type families. So you’ll probably have your own code use functional dependencies too (This was the “solution” I used in my case.) or you have to rewrite the library to use type families. Best wishes, Wolfgang
Henning Günther:
suppose there are two (identical) classes:
class Res a b | a -> b where getRes :: a -> b
and
class Res2 t where type Member t getRes2 :: t -> Member t
It is easy to automatically make every instance of Res2 an instance of res:
instance Res2 a => Res a (Member a) where getRes x = getRes2 x
However, declaring every instance of Res an instance of Res2 seems impossible, as the following doesn't compile
instance Res a b => Res2 a where type Member a = b getRes2 x = getRes x
Question is: How to do this? The reason I need it is because I use a library which uses functional dependencies, but my classes shall be type families.
The last definition is invalid as the right-hand side of a type family instance can only depend on its parameters. However, in type Member a = b you pull 'b' out of thin air. Remember that associated types (ie, type families as part of classes) are only syntactic sugar for sparate type family and class declarations. Obviously, it would be imposible to pull the type instance out of the class in your definition. (The mismatch between FDs and TFs here really is due to FDs being tied to classes and TFs being separable - in that sense TFs are more general than FDs, and hence, you cannot always simulate TFs with FDs.) However, you can wrap an FD library into a TF interface with some additional effort. Using your example for illustration, define the type family and class separately: type family Member a class Res2 a where getRes2 :: a -> Member a Then, implement the catch all class instance as follows: instance Res a (Member a) => Res2 a where getRes2 x = getRes x (This needs -fundecidable-instances. It is perfectly decidable if your FD class is, but GHC doesn't know that.) Now, the additional overhead is that you need to define the type family instances separately; ie, for every class instance of the FD class, such as instance Res Int Bool where getRes x = x == 0 you need to repeat the type mapping: type instance Member Int = Bool I hope this is not too much of a burden in your application. Manuel PS: Hmm, maybe this should go onto the wiki...
participants (3)
-
Henning Günther -
Manuel M T Chakravarty -
Wolfgang Jeltsch