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?
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