[Haskell-cafe] Rank-2 polymorphism and overloading

[Daniel Fischer]

Am Montag 26 April 2010 19:52:23 schrieb Thomas van Noort:
> 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.

Yes, y's type is more general than the type required by f, hence y is an 
acceptable argument for f - even z :: forall a b. a -> b -> Bool is.

> However, it requires y to throw away the provided
> dictionary under the hood, which seems counter intuitive to me.

Why? y doesn't need the dictionary, so it just ignores it.

>
> Regards,
> Thomas