[Haskell-cafe] Rank-2 polymorphism and overloading
daniel.is.fischer at web.de
Mon Apr 26 14:12:57 EDT 2010
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
> 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.
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