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.
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