John J Barton
johnjbarton at johnjbarton.com
Tue Nov 1 10:22:54 PDT 2011
On Tue, Nov 1, 2011 at 9:45 AM, Allen Wirfs-Brock <allen at 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
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...
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