On Sun, Sep 25, 2005 at 11:24:05 -0700, Ashley Winters wrote:
> I can't accept that. While you can infer that $dog will be a Dog at
> that line of code, it isn't being enforced, which means no
> compile-time error. $dog is allowed to store any kind of data, and you
> only know what methods exist in Dog at compile-time. After all, I was
> planning to add a &Dog::as(Cat) method at runtime. Yes, I'm a mad
> scientist. Muahahaha!!!

If you say

my $pet = Dog.new;

if ($condition) {
        $pet = Cat.new;

my Car $spot = $pet;

Then $pet's type is inferred to Cat|Dog, and must be able to contain
either of those, but that is still not a Car.

I'm assuming under use optimize where everything is closed.;

> In order to enforce that level of compile-time type safely, you should
> need to declare my Dog $dog, or stick a pragma up top:

That's the point of my question - why? What do I lose by

 ()  Yuval Kogman <[EMAIL PROTECTED]> 0xEBD27418  perl hacker &
 /\  kung foo master: /me climbs a brick wall with his fingers: neeyah!

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