Implicit 'forall' in data declarations
dave at zednenem.com
Mon Oct 25 04:21:59 EDT 2010
On Mon, Oct 25, 2010 at 3:16 AM, Simon Peyton-Jones
<simonpj at microsoft.com> wrote:
> | 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
> | 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 <dave at zednenem.com>
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