> Now, how transpilers are going to solve Object.mixin super call? 'cause once
> again, that should be solved runtime and I am curious, without caller, how
> transpilers are thinking to solve that.
You can do the following:
SubClass.prototype.foo = function me(x) {
var ssuper = me.homeObject.__proto__;
return 1 + ssuper.foo.call(this, x);
};
Additionally, one would have to make the following assignment for each method m
of SubClass.prototype:
SubClass.prototype.m.homeObject = SubClass.prototype;
Rationale: a method needs to be aware of its (static) position in the prototype
chain if it wants to make a proper super-reference.
More information: http://www.2ality.com/2011/11/super-references.html
--
Dr. Axel Rauschmayer
[email protected]
home: rauschma.de
twitter: twitter.com/rauschma
blog: 2ality.com
_______________________________________________
es-discuss mailing list
[email protected]
https://mail.mozilla.org/listinfo/es-discuss
