On 01.11.2011 21:34, Axel Rauschmayer wrote:
If you have a prototype chain of objects, |this| will always point to
the beginning of the chain (which is also where properties are created
whenever you set a property via |this|). This is what allows methods
in the prototype to access instance properties.
If there is a method m in any of the objects of the chain and it makes
a super-call, you want the search to start in the prototype of m’s
object – and not in the prototype of |this| (which is always the
second object in the prototype chain).
Does that make sense?
No it doesn't. Is there someone on the list who reads it carefully?
Once again. This is a well-known for years issue and has _several_ (at
least three) solutions. One of them (which I call "delete-and-restore
parent link" technique) I showed in the previous letter.
The mentioned technique allows normally to make super calls w/o
hardcoding names of classes. Other techniques (such as using caller and
wrappers) also can be used, but not so efficient.
Dmitry.
On Nov 1, 2011, at 18:22 , John J Barton wrote:
On Tue, Nov 1, 2011 at 9:45 AM, Allen Wirfs-Brock
<al...@wirfs-brock.com <mailto: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
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--
Dr. Axel Rauschmayer
a...@rauschma.de <mailto:a...@rauschma.de>
home: rauschma.de <http://rauschma.de>
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blog: 2ality.com <http://2ality.com>
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