Ben Rudiak-Gould wrote:
Brian Hulley wrote:
I'm puzzled why GHC chooses to create illegal types instead of finding the innermost quantification point ie I would think that
(Ord a=> a->a) -> Int
should then "obviously" be shorthand for
(forall a. Ord a=> a->a) -> Int
and surely this could easily be implemented by just prepending "forall a b c" onto any context restricting a b c... (?)
I agree that it's strange to add an implicit quantifier and then complain that it's in the wrong place. I suspect Simon would change this behavior if you complained about it. My preferred behavior, though, would be to reject any type that has a forall-less type class constraint anywhere but at the very beginning. I don't think it's a good idea to expand implicit quantification. Also, the rule would not be quite as simple as you make it out to be, since
forall a. (forall b. Foo a b => a -> b) -> Int
is a legal type, for example.
This is what I still don't understand: how the above could be a legal type. Surely it introduces 'a' to be anything, and then later retricts 'a' to be related to 'b' via the typeclass 'Foo' ? I would have thought only the following would be legal: f :: (forall a b. Foo a b => a->b) -> Int In other words, in: f :: forall a. (forall b. Foo a b => a->b) -> Int f g = ... how can 'f' pass the dictionary 'Foo a b' to g when 'f' can only choose 'b' but doesn't know anything about 'a'? Where does it get this dictionary from? Regards, Brian.