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Home/ Questions/Q 6616281
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Editorial Team
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Editorial Team
Asked: May 25, 20262026-05-25T20:33:50+00:00 2026-05-25T20:33:50+00:00

I’m not sure why when I overwrite the prototype below with an object(Gadget.prototype =

  • 0

I’m not sure why when I overwrite the prototype below with an object(Gadget.prototype = {
price: 100;}, only new instances of Gadget(theGadget) have access to the new properties.

But when extended(Gadget.prototype.price = 100) All instances have access.

function Gadget(name, color) {
this.name = name;
this.color = color;
this.brand = "Sony";
this.whatAreYou = function(){
return 'I am a ' + this.color + ' ' + this.name;
    }
}

myGadget = new Gadget();

myGadget.brand;


//Gadget.prototype.price = 100;

Gadget.prototype = {
 price: 100,
 rating: 3,

};

myGadget.price;
theGadget = new Gadget();
theGadget.price
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  1. Editorial Team
    Editorial Team
    2026-05-25T20:33:51+00:00Added an answer on May 25, 2026 at 8:33 pm

    It seems pretty obvious to me – each object has a reference to its prototype which is set when the object is first constructed. If you set the prototype to something new:

    Gadget.prototype = {price: 100};
    

    you haven’t changed any references to the old prototype. Only objects created afterwards will have their prototype set to the new value.


    Think of it like the difference between this:

    var a = {foo: true};
    var b = a;
    a = {baz: 'quux'}; // b refers to the original value of a
    

    and this:

    var a = {foo: true};
    var b = a;
    a.baz = 'quux'; // a and b refer to the same object
    
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