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Home/ Questions/Q 8982461
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Editorial Team
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Editorial Team
Asked: June 15, 20262026-06-15T20:35:04+00:00 2026-06-15T20:35:04+00:00

When running this code, the whatever value is first passed in with the constructor

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When running this code, the whatever value is first passed in with the constructor Test(callbacks) becomes the callback that is always called, even in later instantiations of Test

function Test(callbacks) {
    if (callbacks) {
        if (callbacks.callback) {
            this.callback = callbacks.callback;
        }
    }

    this.options.complete = $.proxy(this.options.complete, this);
}

Test.prototype = {
    options: {
        type: "GET",
        complete: function() {
            this.callback();
        }
    },
    callback: function() { console.log("OVERRIDE ME"); },

    execute: function() {
        $.ajax(this.options);
    }
};

var eins = {callback: function() {console.log("AAA");}};
var zwei = {callback: function() {console.log("BBB");}};

var A = new Test(eins);
var B = new Test(zwei);

A.execute();
B.execute();

Running this code, every time you will get the output AAA. How does function() {console.log("AAA");} become a constant value for the prototype?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-15T20:35:05+00:00Added an answer on June 15, 2026 at 8:35 pm

    It all starts with this line:

    this.callback = callbacks.callback;
    

    When you make a call new Test(eins), eins comes in as the callbacks argument. That line then sets this.callback (ie. the “callback” property on your new instance of Test) to the callback property of callbacks, ie. of eins.

    Now, that alone wouldn’t affect B. However, there’s something tricky:

    this.options.complete = $.proxy(this.options.complete, this);
    

    You would think that that would set the “options” property on your Test instance, right? Wrong. The way Javascript works is that if a property isn’t defined on your instance (eg. you didn’t do this.options = something) then Javascript will look up the “prototype chain”, where it will find the prototype’s “options”, and set it (not your instance’s “options”, since your instance doesn’t have one).

    You can fix all this by changing that line to:

    this.options = {complete: $.proxy(this.options.complete, this)};
    

    but of course that would lose your type: "GET", so either you need to do:

    this.options = {type: "GET", complete: $.proxy(this.options.complete, this)};
    

    or you need to base your options off the prototype’s:

    this.options = {};
    for (var key in this.prototype.options) {
        this.options[key] = this.prototype.options[key];
    }
    this.options.complete = $.proxy(this.options.complete, this);
    

    If you happen to be using the (excellent) Underscore library, it even has an extend function for doing this sort of thing more easily:

    this.options = _.extend({}, this.prototype.options,
                            {complete: $.proxy(this.options.complete, this)});
    

    Or (depending on your style preferences):

    this.options = _.extend({}, this.prototype.options);
    this.options.complete = $.proxy(this.options.complete, this);
    

    Incidentally, Underscore also has a _.bind method which is comparable to jQuery’s “proxy”, so you could also do:

    this.options = _.extend({}, this.prototype.options);
    this.options.complete = _.bind(this.options.complete, this);
    
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