I'm fairly new to the concept of JavaScript's prototype concept.
Considering the following code :
var x = function func(){
}
x.prototype.log = function() {
console.log("1");
}
var b = new x();
As I understand it, b.log() should return 1 since x is its prototype. But why is the property b.prototype undefined?
Isn't b.prototype supposed to return the reference to the x function?
Only constructor functions have prototypes. Since x is a constructor function, x has a prototype.
b is not a constructor function. Hence, it does not have a prototype.
If you want to get a reference to the function that constructed b (in this case, x), you can use
b.constructor
The .prototype property of a function is just there to set up inheritance on the new object when the function is invoked as a constructor.
When the new object is created, it gets its internal [[Prototype]] property set to the object that the function's .prototype property points to.
The object itself doesn't get a .prototype property. Its relationship to the object is completely internal.
That's why it works to do b.log(). When the JS engine sees that the b object itself has no log property, it tries to look it up on the objects internal [[Prototype]] object, where it successfully finds it.
To be clear, the [[Prototype]] property is not directly accessible. It's an internal property that is only indirectly mutable via other constructs provided by the JS engine.
All ordinary objects in JavaScript have an internal prototype slot (note: the prototype here does not refer to the prototype property). The ECMAScript standard (http://www.ecma-international.org/ecma-262/6.0/index.html) specifies that this slot is referred to as [[Prototype]]. You could access this slot through the __proto__ property.
__proto__ may not be reliably available across browsers. __proto__ becomes an official property in ECMAScript 6
The prototype property is, however, a property on a constructor function that sets what will become the __proto__ property on the constructed object.
You can access the prototype property of certain types, e.g., the core JavaScript types (Date, Array, and etc). Also a JavaScript function (, which can be regarded as a constructor) has a public prototype property. However, instances of a function do not have a prototype property.
In you case, var b = new x();, b is an instance of function x. Thus b.prototype is undefined. However, b does have an internal [[Prototype]] slot. If you output b.__proto__ in Google Chrome e.g., version 63.0.3239.132, or Firefox such as version 43.0.4
console.log(b.__proto__);
You will see its [[Prototype]] slot as below:
{log: ƒ, constructor: ƒ}
That's it.
And just for your reference, the whole code snippet is put as below:
var x = function() {
};
x.prototype.log = function() {
console.log("1");
}
var b = new x();
b.log(); // 1
console.log(b.prototype); // undefined
console.log(b.__proto__); // {log: ƒ, constructor: ƒ}
console.log(x.prototype); // {log: ƒ, constructor: ƒ}
Before going through your code I want to make sure some concept of prototype that are required to understand your code behavior.
[[prototype]] is a hidden property of a JavaScript object.This hidden property is nothing but a link to Object.prototype(If created by object literals).There is no standard way to access this [[prototype]] property. [[prototype]] property.Here, In case of function this hidden property is a link to Function.prototype.There is also no standard way to access this [[prototype]] property.[[prototype]], Whenever a function object is created,a prototype property is created within it, which is separate from hidden [[prototype]] property.Now coming to your code :
var x = function func(){}
When this line execute , a function object x is created with two links :
x.prototype.log = function() { console.log("1"); }
as we know now that x is a function object so x.prototype is accessible, so here you are able to include log method with it.
var b = new x();
b is an object but not function object .It has that hidden link [[prototype]] but It is not accessible. so when you try to access like b.prototype it gives undefined as a result.If you want to check the prototype of b than you can see (x.prototype).isPrototypeOf(b); it will return true. so you can say that hidden link is referenced to x.prototype.
Here are some facts about prototype :
O is created with O = new func(){} then O[[prototype]] is Function.prototype.O is created with O = {}then O[[prototype]] is Object.prototype.O is created with O = Object.create(obj) then O[[prototype]] is obj.
Because prototype is a property of functions (actually, constructors), since it defines the properties/methods of objects of this class (those which were created from the constructor this prototype belongs). Take a look at this link
