Why does instanceof return false in JavaScript?

Question

In the following code, why does instanceof return false for both Shape and Rectangle? Also why do the own properties of rec include both x and y from the superclass?

    function Shape(x, y) {
        this.x=x;
        this.y=y;
    }
    Shape.prototype.move = function (x, y) {
        this.x += x;
        this.y += y;
        console.log("x = " + this.x + " y = " + this.y);
    };
    function Rectangle(x, y, w, h) {
        Shape.call(this, x, y);
        this.w = w;
        this.h = h;
    }
    Rectangle.prototype = Object.create(Shape.prototype);
    Rectangle.prototype.area = function() {
        return this.w * this.h;
    };
    var rec = new Rectangle(0,0,10,10);
    console.log("instanceof = " + rec instanceof Shape);
    console.log("instanceof = " + rec instanceof Rectangle);
    rec.move(2,3);
    console.log("area = " + rec.area());
    console.log(Object.getOwnPropertyNames(rec));

you should set the prototype for Rectangle to new Shape(). You should check out codeacademy.com if you're new to JavaScript. — Glenn Ferrie
– Glenn Ferrie, Commented Nov 10, 2012 at 23:01
@GlennFerrieLive: Object.create(Shape.prototype) is preferred because it doesn't require an invocation of the Shape constructor, and gives you an empty object. — I Hate Lazy
– I Hate Lazy, Commented Nov 10, 2012 at 23:35

Jonathan Lonowski · Accepted Answer · 2012-11-10 23:05:17Z

9

Because + is evaluated before instanceof. So you're asking if:

"instanceof = " + rec

...a String, is an instanceof your constructors, which it won't be.

Either add parenthesis to force the order:

console.log("instanceof = " + (rec instanceof Shape));

Or, since console.log accepts any number of arguments, pass it as its own:

console.log("instanceof = ", rec instanceof Shape);

edited Nov 10, 2012 at 23:05

answered Nov 10, 2012 at 23:00

Jonathan Lonowski

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