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Home/ Questions/Q 819803
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T02:22:13+00:00 2026-05-15T02:22:13+00:00

I’m trying to extend Object functionality this way: Object.prototype.get_type = function() { if(this.constructor) {

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I’m trying to extend Object functionality this way:

Object.prototype.get_type = function() {
    if(this.constructor) {
        var r = /\W*function\s+([\w\$]+)\(/;
        var match = r.exec(this.constructor.toString());
        return match ? match[1].toLowerCase() : undefined;
    }
    else {
        return typeof this;
    }
}

It’s great, but there is a problem:

var foo = { 'bar' : 'eggs' };
for(var key in foo) {
    alert(key);
}

There’ll be 3 passages of cycle.
Is there any way to avoid this?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-15T02:22:14+00:00Added an answer on May 15, 2026 at 2:22 am

    I, for one, am not completely against extending native types and ECMA-262 5th ed. solves the problems mentioned in other answers and linked articles for us in a nice manner. See these slides for a good overview.

    You can extend any object and define property descriptors that control the behavior of those properties. The property can be made non enumerable meaning when you access the objects properties in a for..in loop, that property will not be included.

    Here’s how you can define a getType method on Object.prototype itself, and make it non enumerable:

    Object.defineProperty(Object.prototype, "getType", {
        enumerable: false,
        writable: false,
        configurable: false,
        value: function() {
            return typeof this;
        }
    });
    
    // only logs "foo"
    for(var name in { "foo": "bar" }) {
        console.log(name); 
    }
    

    The getType function above is mostly useless as it simply returns the typeof object which in most cases will simply be object, but it’s only there for demonstration.

    [].getType();
    {}.getType();
    (6).getType();
    true.getType();
    
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