The following code {-# LANGUAGE FlexibleInstances, TypeFamilies #-} import Control.Applicative class Z t where type W t z :: t -> W t instance Z (a -> b) where type W (a -> b) = a -> b z = id instance Z (IO (a -> b)) where type W (IO (a -> b)) = IO a -> IO b z = (<*>) works fine, but if I try and generalize to from IO to the Applicative classes instance (Applicative m) => Z (m (a -> b)) where type W (m (a -> b)) = m a -> m b z = (<*>) I get the following error Temp.hs:10:9: Conflicting family instance declarations: type instance W (a -> b) -- Defined at Temp.hs:10:9 type instance W (m (a -> b)) -- Defined at Temp.hs:14:9 Failed, modules loaded: none. unless I remove one of the instances, and then it is happy. Is this correct? I don't claim to really understand the rules regarding type classes, but I can't see why these are overlapping. Thanks! -Tyson
Am Donnerstag 04 März 2010 00:17:09 schrieb Tyson Whitehead:
The following code
{-# LANGUAGE FlexibleInstances, TypeFamilies #-}
import Control.Applicative
class Z t where type W t z :: t -> W t
instance Z (a -> b) where type W (a -> b) = a -> b z = id
instance Z (IO (a -> b)) where type W (IO (a -> b)) = IO a -> IO b z = (<*>)
works fine, but if I try and generalize to from IO to the Applicative classes
instance (Applicative m) => Z (m (a -> b)) where type W (m (a -> b)) = m a -> m b z = (<*>)
I get the following error
Temp.hs:10:9: Conflicting family instance declarations: type instance W (a -> b) -- Defined at Temp.hs:10:9 type instance W (m (a -> b)) -- Defined at Temp.hs:14:9 Failed, modules loaded: none.
unless I remove one of the instances, and then it is happy.
Is this correct? I don't claim to really understand the rules regarding type classes, but I can't see why these are overlapping.
Thanks! -Tyson
Because: instance Applicative ((->) a) -- Defined in Control.Applicative so, from the instance Z (a -> b), with b == c -> d, we have an instance Z (a -> (b -> c)) and from instance Z (m (u -> v)), we have, with m == ((->) x), an instance Z (x -> (u -> v))
On March 3, 2010 18:35:26 Daniel Fischer wrote:
Because:
instance Applicative ((->) a) -- Defined in Control.Applicative
so, from the instance Z (a -> b), with b == c -> d, we have an
instance Z (a -> (b -> c))
and from instance Z (m (u -> v)), we have, with m == ((->) x), an
instance Z (x -> (u -> v))
Thanks Daniel, That makes sense. Strangely enough though, I had actually originally tried it with my own Applicative class just in case I was being tripped up by something like the (->) instance you pointed out, and it still didn't work. That is {-# LANGUAGE FlexibleInstances, TypeFamilies #-} newtype I a = I a class A t where ap :: t (a -> b) -> t a -> t b class Z t where type W t z :: t -> W t instance A I where ap (I f) (I x) = I $ f x instance Z (a -> b) where type W (a -> b) = a -> b z = id instance A t => Z (t (a -> b)) where type W (t (a -> b)) = t a -> t b z = ap also gives me Temp.hs:17:9: Conflicting family instance declarations: type instance W (a -> b) -- Defined at Temp.hs:17:9 type instance W (t (a -> b)) -- Defined at Temp.hs:21:9 Failed, modules loaded: none. Is the compiler somehow anticipating that I could add an instance for (->) to A and thus be back to the Applicative class situation? Thanks! -Tyson PS: I asked this here because type classes is a GHC issue, would the haskell- cafe list been a more appropriate place?
Am Donnerstag 04 März 2010 02:39:30 schrieb Tyson Whitehead:
On March 3, 2010 18:35:26 Daniel Fischer wrote:
Because:
instance Applicative ((->) a) -- Defined in Control.Applicative
so, from the instance Z (a -> b), with b == c -> d, we have an
instance Z (a -> (b -> c))
and from instance Z (m (u -> v)), we have, with m == ((->) x), an
instance Z (x -> (u -> v))
Thanks Daniel,
That makes sense. Strangely enough though, I had actually originally tried it with my own Applicative class just in case I was being tripped up by something like the (->) instance you pointed out, and it still didn't work.
Well, GHC takes only the class head into account for instance selection, and u -> (v -> w) matches both, a -> b -- (a == u, b == v -> w) and m (c -> d) -- (m == ((->) u), c == v, d == w), so there's the overlap without any other type classes involved. And since u -> (v -> w) matches both instance heads, type W (u -> (v -> w)) = u -> (v -> w) and type W (((->) u) (v -> w)) = (u -> v) -> (u -> w) are indeed conflicting, so you can't even use OverlappingInstances etc. to make it work.
That is
{-# LANGUAGE FlexibleInstances, TypeFamilies #-}
newtype I a = I a
class A t where ap :: t (a -> b) -> t a -> t b
class Z t where type W t z :: t -> W t
instance A I where ap (I f) (I x) = I $ f x
instance Z (a -> b) where type W (a -> b) = a -> b z = id
instance A t => Z (t (a -> b)) where type W (t (a -> b)) = t a -> t b z = ap
also gives me
Temp.hs:17:9: Conflicting family instance declarations: type instance W (a -> b) -- Defined at Temp.hs:17:9 type instance W (t (a -> b)) -- Defined at Temp.hs:21:9 Failed, modules loaded: none.
Is the compiler somehow anticipating that I could add an instance for (->) to A and thus be back to the Applicative class situation?
The compiler works on an open-world assumption, if the kinds match, there could be an instance defined somewhere.
Thanks! -Tyson
PS: I asked this here because type classes is a GHC issue, would the haskell- cafe list been a more appropriate place?
Either is fine. Cheers, Daniel
On March 3, 2010 21:10:56 Daniel Fischer wrote:
Well, GHC takes only the class head into account for instance selection, and
u -> (v -> w)
matches both,
a -> b -- (a == u, b == v -> w)
and
m (c -> d) -- (m == ((->) u), c == v, d == w),
... <snip> ...
are indeed conflicting, so you can't even use OverlappingInstances etc. to make it work.
Thanks very much for the explanation. I had only read 7.7 (Type families) from the GHC manual. After reading what you wrote and all of 7.6 (Class and instances declarations) a couple of times I think I've got it. As it says in 7.7.2.2.2, "[t]he instance declarations of a type family used in a single program may only overlap if the right-hand sides of the overlapping instances coincide for the overlapping types." Cheers! -Tyson
participants (2)
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Daniel Fischer -
Tyson Whitehead