On Tue, Aug 2, 2011 at 5:57 AM, Patrick Browne
data Eq a => Set1 a = NilSet1 | ConsSet1 a (Set1 a) data Set2 a = NilSet2 | ConsSet2 a (Set2 a) deriving Eq
The former declaration is a language wart, IMO. All it does is attach a restriction to the constructors of Set1; try
:t NilSet1 NilSet1 :: Eq a => Set1 a :t NilSet2 NilSet2 :: Set2 a
But it doesn't give you that restriction back when you use it:
let f (ConsSet1 v _) = v == v :t f f :: Eq a => Set1 a -> Bool
You'd think that since you had to provide the evidence (Eq a) when you constructed ConsSet1, that it'd be available, but it's not. -- Seems to have same type
:t ConsSet2 1 NilSet2 ConsSet2 1 NilSet2 :: forall t. (Num t) => Set2 t :t ConsSet1 1 NilSet1 ConsSet1 1 NilSet1 :: forall t. (Num t) => Set1 t
Remember that Eq is a superclass of Num.
let f1 x = ConsSet1 x NilSet1 let f2 x = ConsSet2 x NilSet2 :t f1 f1 :: Eq a => a -> Set1 a f2 :: a -> Set2 a
'deriving Eq' on the definition of Set2 makes the instance 'instance Eq a => Eq (Set2 a)'. So you can construct Set2 at any type, and when the underlying type 'a' has equality, then 'Set2 a' does as well. -- ryan