JavaScript: inheritance

Question

I tried to make inheritance, but I didn't expect that this.array will act like static member. How can I make it 'protected/public':

function A() {
    this.array = [];
}

function B() {
    this.array.push(1);
}
B.prototype.constructor=B;
B.prototype = new A();

Firebug:

>>> b = new B();
A { array=[1]}
>>> b = new B();
A { array=[2]}
>>> b = new B()
A { array=[3]}

Your last two lines are backwards, by the way. Set prototype.constructor after you set prototype. — Ry-
– Ry- ♦, Commented Jan 26, 2012 at 2:33
I realize this is just an example, but why don't you do this.array = [1] instead? — Ry-
– Ry- ♦, Commented Jan 26, 2012 at 2:35
Original code looks like: function Controller(uri) { this._controllerURI = uri; this._controllers = []; this._views = []; this._events = []; }. It's just a sample. — DraganS
– DraganS, Commented Jan 26, 2012 at 2:45
This question is similar to: How should the `.prototype` property be set up in a derived old-style constructor?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — dumbass
– dumbass, Commented Dec 26, 2024 at 22:23

user1170510 · Accepted Answer · 2012-01-26 02:34:58Z

2

Not "private/protected", but this will make a new Array for each B.

function A() {
    this.array = [];
}

function B() {
    A.apply(this); // apply the A constructor to the new object
    this.array.push(1);
}
// B.prototype.constructor=B; // pointless
B.prototype = new A();

answered Jan 26, 2012 at 2:34

user1170510

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2 Comments

Ry- Over a year ago

I think you mean A.call(this);.

DraganS Over a year ago

@user(\d)* Perfect. Thanks. @minitech: could be apply - it works. Where I can find docs/sample code for your $Query replacement?

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