在没有.prototype的情况下向构造函数添加新属性

Question

when I have function I want to use as a constructor, say:

function clog(x){
    var text = x;
    return console.log(text );
}

And I have made some instances of it

var bla = new clog();

Now I want to add new functionality it, so I would use

clog.prototype.alert = alert(text);

what would be the difference if I simply do:

clog.alert = alert(text);

Would that not be inherited by the objects that clog is their prototype?

Answer 1

Instances created by a constructor function ( clog in your case) inherit a reference to the clog.prototype object. So if you add a property to clog.prototype , it will show up on instances. If you add a property to clog itself, it will not show up on instances.

There are some issues with your quoted code, so let's look at an abstract example:

function Foo() {
}
Foo.prototype.bar = "I'm bar on Foo.prototype";
Foo.bar = "I'm bar on Foo";

var f = new Foo();
console.log(f.bar); // "I'm bar on Foo.prototype"
// E.g., `f` inherits from `Foo.prototype`, not `Foo`

// And this link is live, so:
Foo.prototype.charlie = "I'm charlie on Foo.prototype";
console.log(f.charlie); // "I'm charlie on Foo.prototype";

From your comment below:

I don't understand why new properties added directly to Foo would be ignored by the prototype chain?

Because it's Foo.prototype , not Foo , that is the prototype for objects created via new Foo() .

isn't prototype simply points to the constructor object?

No, Foo and Foo.prototype are completely distinct objects. Foo is a function object, which like all function objects can have properties. One of Foo 's properties is prototype , which is a non-function object that's initially blank other than a constructor property that points back to Foo . It's Foo.prototype , not Foo , that instances created via new Foo get as their prototype. Foo 's only role is to create objects that use Foo.prototype as their prototype. (Actually, in Foo 's case, it just initializes those objects; they're created by the new operator. With a traditional function like Foo , new creates the object. If this code were using ES2015+ class syntax, new wouldn't create the object, it would leave that to Foo [if Foo were a base class constructor] or Foo 's ultimate base class [if Foo were a subclass constructor].)

If I do Foo.newProp = "new addition" why is f.newProp => undefined ?

(To avoid confusion, I've changed Foo.new = ... to Foo.newProp = ... above, since new is a keyword. While you can use it like you did as of ES5, it's best not to.)

Because Foo.newProp has virtually nothing to do with f . You can find it, on f.constructor.newProp , since f.constructor is Foo .

Some ASCII-art:

Given this code:

function Foo() {
}
Foo.prototype.bar = "I'm bar on Foo.prototype";
Foo.bar = "I'm bar on Foo";

we have these objects with these properties (some omitted for clarity):

+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
        |                                       |
        V                 +−−−−−−−−−−−−−−−−−−+  |
+−−−−−−−−−−−−−−−−+    +−−>| [String]         |  |
| Foo [Function] |    |   +−−−−−−−−−−−−−−−−−−+  |
+−−−−−−−−−−−−−−−−+    |   | "I'm bar on Foo" |  |
| bar            |−−−−+   +−−−−−−−−−−−−−−−−−−+  |
| prototype      |−−−−+                         |
+−−−−−−−−−−−−−−−−+    |                         |
                      +−−−−−−−−−−+              |
                                 |              |
                                 V              |
                               +−−−−−−−−−−−−−+  |
                               | [Object]    |  |
                               +−−−−−−−−−−−−−+  |
                               | constructor |−−+   +−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
                               | bar         |−−−−−>| [String]                   |
                               +−−−−−−−−−−−−−+      +−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
                                                    | "I'm bar on Foo.prototype" |
                                                    +−−−−−−−−−−−−−−−−−−−−−−−−−−−−+

Now if we do

var f = new Foo();

we have (new stuff in bold ):

+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
        |                                          |
        V                 +−−−−−−−−−−−−−−−−−−+     |
+−−−−−−−−−−−−−−−−+    +−−>| [String]         |     |
| Foo [Function] |    |   +−−−−−−−−−−−−−−−−−−+     |
+−−−−−−−−−−−−−−−−+    |   | "I'm bar on Foo" |     |
| bar            |−−−−+   +−−−−−−−−−−−−−−−−−−+     |
| prototype      |−−−−+                            |
+−−−−−−−−−−−−−−−−+    |                            |
                      +−−−−−−−−−−−−−+              |
                                    |              |
                                    V              |
+−−−−−−−−−−−−−−−+                 +−−−−−−−−−−−−−+  |
| f [Object]    |          +−−−−−>| [Object]    |  |
+−−−−−−−−−−−−−−−+          |      +−−−−−−−−−−−−−+  |
| [[Prototype]] |−−−−−−−−−−+      | constructor |−−+  +−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
+−−−−−−−−−−−−−−−+                 | bar         |−−−−>| [String]                   |
                                  +−−−−−−−−−−−−−+     +−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
                                                      | "I'm bar on Foo.prototype" |
                                                      +−−−−−−−−−−−−−−−−−−−−−−−−−−−−+

([[Prototype]] is an object's internal field referring to its prototype. This is accessible via Object.getPrototypeOf [or __proto__ on JavaScript engines on web browsers, but don't use __proto__ , it's just for backward compatibility with old SpiderMonkey-specific code.)

Now suppose we do this:

f.charlie = "I'm charlie on f";

All that changes is the f object (new stuff in bold ):

+−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
        |                                          |
        V                 +−−−−−−−−−−−−−−−−−−+     |
+−−−−−−−−−−−−−−−−+    +−−>| [String]         |     |
| Foo [Function] |    |   +−−−−−−−−−−−−−−−−−−+     |
+−−−−−−−−−−−−−−−−+    |   | "I'm bar on Foo" |     |
| bar            |−−−−+   +−−−−−−−−−−−−−−−−−−+     |
| prototype      |−−−−+                            |
+−−−−−−−−−−−−−−−−+    |                            |
                      +−−−−−−−−−−−−−+              |
                                    |              |
                                    V              |
+−−−−−−−−−−−−−−−+                 +−−−−−−−−−−−−−+  |
| f [Object]    |          +−−−−−>| [Object]    |  |
+−−−−−−−−−−−−−−−+          |      +−−−−−−−−−−−−−+  |
| [[Prototype]] |−−−−−−−−−−+      | constructor |−−+  +−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
| charlie       |−−−−−−−−−−+      | bar        |−−−−−>| [String]                   |
+−−−−−−−−−−−−−−−+          |      +−−−−−−−−−−−−−+     +−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
                           |                          | "I'm bar on Foo.prototype" |
                           |                          +−−−−−−−−−−−−−−−−−−−−−−−−−−−−+
                           |
                           |      +−−−−−−−−−−−−−−−−−−−−+
                           +−−−−−>| [String]           |
                                  +−−−−−−−−−−−−−−−−−−−−+
                                  | "I'm charlie on f" |
                                  +−−−−−−−−−−−−−−−−−−−−+

f now has its own property, called charlie . This means that these two statements:

console.log(f.charlie); // "I'm charlie on f"
console.log(f.bar);     // "I'm bar on Foo.prototype"

Get processed slightly differently.

Let's look at f.charlie first. Here's what the engine does with f.charlie :

1. Does f have its own property called "charlie" ?
2. Yes; use the value of that property.

Simple enough. Now let's look at how the engine handles f.bar :

1. Does f have its own property called "bar" ?
2. No; does f have a prototype?
3. Yes; does f 's prototype have a property called "bar" ?
4. Yes; use the value of that property.

So there's a big difference between f.charlie and f.bar : f has its own property called charlie , but an inherited property called bar . And if f 's prototype object hadn't had a property called bar , its prototype object (in this case, Object.prototype ) would be checked, and so on up the chain until we run out of prototypes.

You can test whether a property is an "own" property, btw, using the hasOwnProperty function that all objects have:

console.log(f.hasOwnProperty("charlie")); // true
console.log(f.hasOwnProperty("bar"));     // false

Answering your question from the comments:

I make function Person(first_name, last_name) {this.first_name = first_name; this.last_name = last_name;} function Person(first_name, last_name) {this.first_name = first_name; this.last_name = last_name;} and then var ilya = new Person('ilya', 'D') how does it resolves the inner name properties?

Within the call to Person that's part of the new Person(...) expression, this refers to the newly-generated object that will be returned by the new expression. So when you do this.prop = "value"; , you're putting a property directly on that object, nothing to do with the prototype.

Putting that another way, these two examples result in exactly the same p object:

// Example 1:
function Person(name) {
    this.name = name;
}
var p = new Person("Fred");

// Example 2:
function Person() {
}
var p = new Person();
p.name = "Fred";

---

Here are the issues with the quoted code I mentioned:

Issue 1: Returning something from a constructor function:

function clog(x){
    var text = x;
    return console.log(text ); // <=== here
}

99.9999% of the time, you don't want to return anything out of a constructor function. The way the new operation works is:

1. A new blank object is created.
2. It gets assigned a prototype from the constructor's prototype property.
3. The constructor is called such that this refers to the new object.
4. If the constructor doesn't return anything, or returns something other than an object , the result of the new expression is the object created in step 1.
5. If the constructor function returns an object, the result of the new operation is that object instead.

So in your case, since console.log doesn't return anything, you just remove the return keyword from your code. But if you used that return xyz(); construct with a function that returned an object, you'd mess up your constructor function.

Issue 2: Calling functions rather than referring to them

In this code:

clog.prototype.alert = alert(text);

you're calling the alert function and assigning the result of it to a property called alert on clog.prototype . Since alert doesn't return anything, it's exactly equivalent to:

alert(text);
clog.prototype.alert = undefined;

...which probably isn't what you meant. Perhaps:

clog.prototype.alert = function(text) {
    alert(text);
};

There we create a function and assign a reference to it to the alert property on the prototype. When the function is called, it will call the standard alert .

Issue 3: Constructor functions should be initially capped

This is just style, but it's overwhelmingly standard: Constructor functions (functions meant to be used with new ) should start with an upper case letter, so Clog rather than clog . Again, though, this is just style.

Answer 2

Adding a clog.alert function would simply a attach static function to the clog object. It will not be inherited and will not have access to the instance created with new clog(); in the alert function.

Adding clog.prototype.alert will make the new clog(); object you create inherit the function, and you will also have access to the instance inside using the this keyword.

function John() {
    this.id = 1;
}

John.doe = function() {
  console.log(this);
  console.log(this.id); // undefined
}

John.prototype.doe = function() {
    console.log(this);
};

John.doe(); // the John object

var me = new John();
me.doe(); // the instance, inherited from prototype
console.log(me.id); // 1

Answer 3

Any property added to the constructor will act as a static property that can only be accessed by referring to the constructor object (ie, function) and not using any instance object of it. Its like a Class property and not an instance property.