object in prototype is inherited as reference

Question

I want to inherit new object instance using prototype.

Test case:

var MyObj = function() {}
MyObj.prototype.objName = {} 
// I want this to be a differen         


        

Answer 1

This means that objects in prototype are not inherited as its copies but instead as its reference.

Just to be clear. First of all in JavaScript all objects are passed by reference, not by value. Only primitives are passed by value. Second, you're not actually "copying" or "passing" anything. When you set a prototype, you're creating a prototype's chain. It means that in your case:

var MyObj = function() {};
MyObj.prototype.objName = {} ;

var o1 = new MyObj ();
var o2 = new MyObj ();

Both o1 and o2 doesn't have any property called objName, and you can simply test it with:

console.log(Object.keys(o1)); // []

When JS see a code like o1.objName, as first thing checks if the object has this property, and if it has, use it. If not, start to looking in the prototype's chain, starting by the prototype of o1, that is MyObj.prototype: it found the properties objName, and returns it. If it didn't find it, then JS will continue to check the prototype of MyObj.prototype, and so on. So, here the point: MyObj.prototype it's an object: and you shared that object between o1 and o2. That's why the instance of objName is the same. It's exactly the same logic of having:

function objName(obj) {
    return "objName" in obj ? obj.objName : O.objName;
}

var O = { objName: [] };
var foo = {};
var bar = {};

objName(foo).push(0);
objName(bar).push(1);

So, you can't put in prototype any object that is not meant to be shared across the objects creates using that prototype. I would say that shared states like that is also a bad practice that should be avoided, that's why in general prototype shouldn't have such property. It's still not clear to me why you can't modify the constructor, but the point is: when you create the instance of your object, you have to "setup" it. Usually, calling the constructor, but any function is fine. This is made also when you want to support inheritance, and calling the "super" constructor to initialize your object.