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?