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 _______________________________________________ Haskell-Cafe mailing list [email protected] http://www.haskell.org/mailman/listinfo/haskell-cafe
