Thomas van Noort wrote:
Hello all,
I'm having difficulties understanding rank-2 polymorphism in combination with overloading. Consider the following contrived definition:
f :: (forall a . Eq a => a -> a -> Bool) -> Bool f eq = eq True True
Then, we pass f both an overloaded function and a regular polymorphic function:
x :: forall a . Eq => a -> a -> Bool x = \x y -> x == y
y :: forall a . a -> a -> Bool y = \x y -> True
g :: (Bool, Bool) g = (f x, f y)
Could someone explain to me, or point me to some reading material, why g is correctly typed?
I understand that x's type is what f expects, but why does y's polymorphic type fulfill the overloaded type of f's argument? I can imagine that it is justified since f's argument type is more restrictive than y's type. However, it requires y to throw away the provided dictionary under the hood, which seems counter intuitive to me.
f requires a function that is able to compute, for two values of type a (which instantiates Eq), a Boolean. y certainly fulfills that requirement: it does not even require that the values are of a type instantiating Eq. This is also well-typed and might or might not be enlightening:
z :: forall a. Eq a => a -> a -> Bool z = y
g :: (Bool, Bool) g = (f x, f z) -- note the use of z here instead of y
Jochem -- Jochem Berndsen | jochem@functor.nl