On Tue, Nov 1, 2011 at 9:45 AM, Allen Wirfs-Brock <al...@wirfs-brock.com> wrote: > > On Nov 1, 2011, at 6:53 AM, Jeremy Ashkenas wrote:
> The complication of super is that each "super call" requires two independent > pieces of state in additional to the method arguments: the object that will > be used as the |this| value of the resulting method invocation and the > object where property lookup will begin when searching for the method. > Let's call this the "lookup-point". |super| when used to access a method > property really represents a pair of values (this,lookup-point). In the > general case, these are not the same value so they must be independently > captured and represented. > Where do these two pieces of state come from? The |this| value of a super > call is simply the |this| that was dynamically passed into the calling > method. Where does the look-up point come from. There are fundamentally > two possibilities: > a) it is dynamically passed to every method invocation, ;just like |this| > currently is > b) it is statically associated with the method in some way. I'm just curious and I think understanding the following point would help developers: Why isn't the |super| lookup-point |this.getPrototypeOf()| (The protolink or [[Prototype]])? I thought that if let f = new F(); and F inherits from G, then f.foo() will have this.getPrototypeOf() === G and that is what super would be? Somewhere I went wrong... jjb _______________________________________________ es-discuss mailing list es-discuss@mozilla.org https://mail.mozilla.org/listinfo/es-discuss
