Terminology (created by me, but I think it explains well what is going on):
- |this| points to the object where property lookup starts. It always points to the beginning of the prototype chain. - |here| points to the object where a property was found. |here| can point to any object in the prototype chain. |super| starts property lookup in the prototype of |here|, but does not change |this|. That is, a method invoked via |super| still has the same |this| and property lookup via |this| is unchanged. If the super-method again uses |super|, then property lookup will begin in the prototype of |here| (= where the super-method has been found). Etc. > From: Mariusz Nowak <[email protected]> > Date: June 20, 2011 13:49:20 GMT+02:00 > Subject: Re: Making "super" work outside a literal? > > It's most sane proposal I think. However few things are not obvious to me, > will following evaluate as I assume: > > var A = { > one: function () { > return 'A.foo'; > } > }; > var B = Object.create(A); > > var C = Object.create(B); > C.one = function () { > return super.one(); > }; > > var c1 = Object.create(C); > obj.one(); // 'A.foo' That would be c1.one(), right? |here| === C and thus the search for super.one starts in B and finds A.one. > B.two = function () { > this.three(); > }; > B.three = function () { > return 'B.three'; > }; > C.two = function () { > super.two(); > }; > C.three = function () { > return 'C.three'; > }; > > var c2 = Object.create(C); > c2.two(); // C.three |here| === C and thus super.two === B.two B.two() uses the original and unchanged |this| which is still c2. => this.three === C.three SUMMARY: I agree with your findings. -- Dr. Axel Rauschmayer [email protected] twitter.com/rauschma home: rauschma.de blog: 2ality.com
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