On Fri, Mar 8, 2013 at 2:46 PM, Scott Miles wrote:
> I also want to keep ES6 classes in mind. Presumably in declarative form I
> declare my class as if it extends nothing. Will 'super' still work in that
> case?
>
If you extend nothing (null) as in:
class Foo extends null {
m() {
super();
I also want to keep ES6 classes in mind. Presumably in declarative form I
declare my class as if it extends nothing. Will 'super' still work in that
case?
Scott
On Fri, Mar 8, 2013 at 11:40 AM, Scott Miles wrote:
> Mostly it's cognitive dissonance. It will be easy to trip over the fact
> that
Mostly it's cognitive dissonance. It will be easy to trip over the fact
that both things involve a user-supplied prototype, but they are required
to be critically different objects.
Also it's hard for me to justify why this difference should exist. If the
idea is that element provides extra conven
If you have a tag name it is easy to get the prototype.
var tmp = elementElement.ownerDocument.createElement(tagName);
var prototype = Object.getPrototypeOf(tmp);
On Fri, Mar 8, 2013 at 12:16 PM, Dimitri Glazkov wrote:
> On Thu, Mar 7, 2013 at 2:35 PM, Scott Miles wrote:
>> Currently, if I docu
On Thu, Mar 7, 2013 at 2:35 PM, Scott Miles wrote:
> Currently, if I document.register something, it's my job to supply a
> complete prototype.
>
> For HTMLElementElement on the other hand, I supply a tag name to extend, and
> the prototype containing the extensions, and the system works out the
>
Currently, if I document.register something, it's my job to supply a
complete prototype.
For HTMLElementElement on the other hand, I supply a tag name to extend,
and the prototype containing the extensions, and the system works out the
complete prototype.
However, this ability of HTMLElementEleme
