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  • Super-references

Super-references

# Axel Rauschmayer (13 years ago)

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: www.2ality.com/2011/11/super-references.html

# Andrea Giammarchi (13 years ago)

does that work runtime with more than one super ?

# Andrea Giammarchi (13 years ago)

let me reformulate .. do you realize Object,mixing can be done runtime, so not possible to predefine the super, and that you method fails if you have more than a super in such form ? If not, can you write B extends A and C extends B and use super in both C and B methods going up to A or do what my example was doing ?

Even easier, something like this:

function A() { this.test('Hello!'); } A.prototype.test = function (what) { alert(what); };

function B() { this.super(); } B.prototype = poo.inherit(A.prototype); B.prototype.constructor = B; B.prototype.test = function (what) { this.super(what); };

poo.superable(B.prototype);

new B; // will alert Hello! dong these steps: // 1. invokes B.prototype.constructor which // 2. invokes A.prototype.constructor which // 3. invokes B.prototype.test which // 4. invokes A.prototype.test

# Axel Rauschmayer (13 years ago)

let me reformulate .. do you realize Object,mixing can be done runtime, so not possible to predefine the super, and that you method fails if you have more than a super in such form ? If not, can you write B extends A and C extends B and use super in both C and B methods going up to A or do what my example was doing ?

You don’t predefine super, you only have to keep [[HomeObject]] updated. That’s what Object.mixin does. Then [[HomeObject]] (or rather, its prototype) is the starting point for looking for super-properties.

Even easier, something like this:

function A() { this.test('Hello!'); } A.prototype.test = function (what) { alert(what); };

function B() { this.super(); } B.prototype = poo.inherit(A.prototype); B.prototype.constructor = B; B.prototype.test = function (what) { this.super(what); };

poo.superable(B.prototype);

new B; // will alert Hello! dong these steps: // 1. invokes B.prototype.constructor which // 2. invokes A.prototype.constructor which // 3. invokes B.prototype.test which // 4. invokes A.prototype.test

Yes, that will work. But you can’t find super via this (which changes dynamically), you must start your search in the prototype of the object in which a method is stored.

# Andrea Giammarchi (13 years ago)

I CAN find super via this, the qwhole point about caller since this morning was about the fact I could do stuff I cannot do anymore.

That said, in 2010: webreflection.blogspot.com/2010/01/javascript-super-bullshit.html

late 2009 wrote early 2010: webreflection.blogspot.com/2010/01/better-javascript-classes.html

And benchmark about all this in February 2009: webreflection.blogspot.com/2009/02/on-javascript-inheritance-performance.html

So, just in case, have an extra look too on what developers wrote before your blog did, thank you.

Thanks for your examples in any case.

# Axel Rauschmayer (13 years ago)

The question I tried to answer was:

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.

I assumed you didn’t know, which is why I wrote an answer.