On Mon, Oct 25, 2010 at 3:16 AM, Simon Peyton-Jones
| On a related note, these are also apparently allowed (in 6.10.4): | f :: forall a. (Eq a => a -> a) -> a -> a | -- the Eq context prevents the function from ever being called.
That's not true. E.g. f ((==) True) True works fine.
What I meant is that f cannot call its argument. That is, f :: forall a. (Eq a => a -> a) -> a -> a f g x = g x is ill-typed.
| g :: forall a. Ord a => (Eq a => a -> a) -> a -> a | -- the Eq context is effectively ignored
That's a bit more true, because Ord a implies Eq a, but still not really. It still says that you must pass evidence for equality to g's argument.
Is that different from forall a. Ord a => (a -> a) -> a -> a?
--
Dave Menendez