I've been reading about the prototype chain in JavaScript and came to two slightly different definitions. It is said that every object in JavaScript has a prototype and that prototype in turn has another prototype.
The top prototype (Grand) may also have prototype and the chain can continue. Now the chain will stop at one last object. JavaScript: The Good Parts says the chain terminates at Object.prototype
and MDN says null
is the final link where the chain terminates.
Javascript: The Good Parts
Every object is linked to a prototype object from which it can inherit properties. All objects created from object literals are linked to
Object.prototype
, an object that comes standard with JavaScript.
MDN
Each object has an internal link to another object called its prototype. That prototype object has a prototype of its own, and so on until an object is reached with null as its prototype. null, by definition, has no prototype, and acts as the final link in this prototype chain.
null
and Object.prototype
one and the same thing?Object.prototype
?var x = { len: 4, breadth: 5}
. Would JavaScript automatically create its prototype x.prototype
? And how long would the prototype chain be? Would x.prototype
have only one prototype, Object.prototype
, making a 3 point chain?It is like, if New York has an envelope and inside, it says Colorado, and Colorado has an envelope, and inside, it says San Francisco, and San Francisco has an envelope, and inside, it says "none". So is San Francisco end of the chain, or is "none" the end of chain? It may depend on how you look at it. But one thing is for sure: it points up and up the chain, for inheritance purpose (prototypal inheritance), until it reaches null
, which means can't go further up. And make sure you know that, to go up and up the chain, it is __proto__
. It is not prototype
.
Object.prototype
, Animal.prototype
are different from x.__proto__
. The former are function objects (Object, Animal) having a prototype
property pointing to a prototype object. And x.__proto__
is how the prototype chain is followed upward. To go up and up, it is x.__proto__.__proto__
and so on. See JavaScript's Pseudo Classical Inheritance diagram to understand it more.
Object.prototype
refers to a prototype object. Quoted from MDN, null
"represents the intentional absence of any object value. It is one of JavaScript's primitive values." So Object.prototype
and null
are not the same thing.
All JavaScript objects will have obj.__proto__
referring ultimately to what Object.prototype
refers to. If it is not obj.__proto__
, then it is obj.__proto__.__proto__
. If not, just go up, and up, and it will reach the prototype object which Object.prototype
refers to. And at this point, when you go up one level (by adding a .__proto__
, then you get null
. You can try it in Google Chrome's developer's tool:
x = { a : 1 }
> Object {a: 1}
x.__proto__ === Object.prototype
> true
x.__proto__.__proto__
> null
Object.prototype.__proto__
> null
What is the end of prototype chain in JavaScript ...?
Null. The only authority on the language is ECMA-262.
Are objects, which are not created from object literals, not linked to
Object.prototype
?
They may or many not be, e.g.
var x = Object.create(null)
has a [[Prototype]]
of null, whereas:
var y = {};
has a [[Prototype]]
of Object.prototype.
Say I have an object
var x = { len: 4, breadth: 5}
. Would JavaScript automatically create its prototypex.prototype
?
No. Function objects have default prototype objects. Plain objects have a default [[Prototype]]
(i.e. internal prototype property) that is Object.prototype (unless constructed as above).
And how long would the prototype chain be? Would
x.prototype
have only one prototype,Object.prototype
, making a 3 point chain?
Two values: Object.prototype and null.
How does JavaScript internally create automatic prototypes?
However it likes, the language specification does not define implementation, only behaviour.
What is the end of prototype chain in javascript --
null
orObject.prototype
? Or are null and Object.prototype one and the same thing?
null
. Consider this code in a normal JavaScript environment:
var o = {};
console.log(o.i_am_a_property_that_does_not_exist);
That property accessor operation (o.i_am_a_property_that_does_not_exist
) ends up going to the OrdinaryGet abstract operation defined by the specification with O set to the o
object above and P set to "i_am_a_property_that_does_not_exist"
. That operation starts like this:
a. Let parent be ? O.[GetPrototypeOf].
b. If parent is null, return undefined.
c. Return ? parent.[[Get]](P, Receiver).
For my example above, that [[Get]] operation in 3.c. ends up calling OrdinaryGet recursively until we run out of prototypes. As we can see, the chain ends when we reach null
.
Moreover, it's entirely possible to have an object with a prototype chain that doesn't include Object.prototype
at all (we'll see some in a moment). So clearly Object.prototype
can't be the end of the prototype chain.
Are objects, which are not created from object literals, not linked to
Object.prototype
?
The vast majority will be linked to Object.prototype
directly or indirectly. Consider:
function Thing() {
}
var t = new Thing();
t
's prototype is the object referenced by Thing.prototype
. Thing.prototype
's prototype is Object.prototype
. So t
is linked to Object.prototype
(indirectly).
But it's entirely possible for an object not to be linked to Object.prototype
. Here's one way:
var o = Object.create(null);
Object.create
lets us create an object with the prototype we specify in the first argument (null
in the above). So o
above has no prototype, its [[Prototype]] internal slot (where objects remember their prototypes) is null
.
Here's another:
function Thing() {
}
Thing.prototype = Object.create(null);
var t = new Thing();
In that case, although t
has a prototype, its prototype's prototype is null; t
isn't linked to Object.prototype
at all.
Say I have an object
var x = { len: 4, breadth: 5}
. Would JavaScript automatically create it's prototypex.prototype
.
An object's prototype is not a property called prototype
, so no, the JavaScript engine wouldn't create x.prototype
. An object's prototype is linked via its [[Prototype]] internal slot, which is not directly observable but can be retrieved via Object.getPrototypeOf
. (There's also the legacy __proto__
property, but it's best not to use it, not least because A) It's an optional part of the specification, and B) Not all objects inherit from Object.prototype
, which is where __proto__
is defined.)
The prototype
property is just used on functions to determine what object to use as the [[Prototype]] of new objects created via new
with that function. Non-function objects don't have it, and if they did, it wouldn't be any more special than a property called foo
or bazinga
.
And how long would the prototype chain be? Would
x.prototype
have only one prototypeObject.prototype
making a 3 point chain?
You're close, but again, the prototype
property is not the prototype of the object, and non-function objects typically won't have a prototype
property. For var x = { len: 4, breadth: 5}
, the inheritance chain would be:
x
x
's [[Prototype]] (which is Object.prototype
)Object.prototype
's [[Prototype]], which is null
So quite short; 1, 2, or 3 depending on whether you want to count x
and whether you want to count null
.
How does JavaScript internally creates automatic prototypes?
It doesn't, other than the ones defined by the spec (e.g., Object.prototype
and such). The closest it comes is that for all function
functions, the JavaScript engine automatically creates an object and assigns that object to the function's prototype
property just in case that function is used as a constructor (via new
). (It doesn't do this with arrow functions or generators.)
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