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Home/ Questions/Q 6383329
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T02:38:10+00:00 2026-05-25T02:38:10+00:00

I’ve created a superclass Control and a subclass LED . Control ‘s initializer sets

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I’ve created a superclass Control and a subclass LED. Control‘s initializer sets several attributes, like height, width, that sort of stuf. However, when I create an instance of LED – with the the proper data to fill the attributes that it should inherit – the inherited attributes don’t show up when logging the instance.

Here are the two constructors:

var Control = Class.create({
    initialize: function(control){
        log('info','Beginning creation of control using following control object.');
        log('log',control);

        // Define self to prevent conflicts with frameworks
        if(!self){ var self = this; }

        this.type = null;
        this.size = { w: control.w, h: control.h };
        this.position = { x: control.x, y: control.y };
        this.color = control.color;
        this.name = control.name;
        this.active = false;
    }
});

var LED = Class.create(Control,{
    initialize: function($super){
        $super;
        this.type = "led";
    }
});

I’ve created a small Heroku site to demonstrate: http://blazing-dusk-603.heroku.com/interfaces/16, if you open the console you’ll see a ton of logs, the one at the bottom is the instance of LED, without a size, position, color, name or active attribute.

This is the first time I use prototypejs to inherit so maybe I’ve misunderstood something, help would be appreciated, of course;)

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-25T02:38:10+00:00Added an answer on May 25, 2026 at 2:38 am

    If you want to call the superclass method, you probably should actually do so:

      $super.call(this, this); // not exactly sure what "$super" is supposed to be
      $super.initialize(this); // maybe?
    

    (or something like that). In any case, just mentioning the variable like your code does will have no effect.

    edit — according to the Prototype docs, “$super” is a reference to the parent class function, so if you want to call it you’ll have to pass in a “control” parameter. I don’t know what that’s supposed to be, but your base class “initialize()” clearly wants it to be something non-null.

    Really, it looks to me as if your subclass “initialize()” isn’t really following the API pattern of the parent class:

    var LED = Class.create(Control, {
      initialize: function($super, control) {
        $super.call(this, control);
    

    seems like it’s closer to what you want.

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