Asked  7 Months ago    Answers:  5   Viewed   18 times

Trying to bend by head around Javascript's take on OO...and, like many others, running into confusion about the constructor property. In particular, the significance of the constructor property, as I can't seem to make it have any effect. E.g.:

function Foo(age) {
    this.age = age;
}

function Bar() {
    Foo.call(this, 42);
    this.name = "baz"; 
}

Bar.prototype = Object.create(Foo.prototype); 
var b = new Bar;    

alert(b.constructor); // "Foo". That's OK because we inherit `Foo`'s prototype.
alert(b.name);        // "baz". Shows that Bar() was called as constructor.
alert(b.age);         // "42", inherited from `Foo`.

In the above example, the object b seems to have had the right constructor called (Bar) – and it inherits the age property from Foo. So why do many people suggest this as a necessary step:

Bar.prototype.constructor = Bar;

Clearly, the right Bar constructor was called when constructing b, so what impact does this prototype property have? I am curious to know what practical difference it actually makes to have the constructor property set 'correctly'—as I can't see it having any affect on which constructor is actually called after an object is created.

 Answers

25

September 2020 Update

The answer below is from the days of ECMAScript 3 and the first sentence is no longer true because since ECMAScript 6, the constructor property is used in a few places. However, I think the overall gist still applies. Thanks to T. J. Crowder for pointing that out in the comments, and please read his answer for a fuller picture of the current situation.

Original answer

The constructor property makes absolutely no practical difference to anything internally. It's only any use if your code explicitly uses it. For example, you may decide you need each of your objects to have a reference to the actual constructor function that created it; if so, you'll need to set the constructor property explicitly when you set up inheritance by assigning an object to a constructor function's prototype property, as in your example.

Tuesday, June 1, 2021
 
Strae
answered 7 Months ago
82

Bind creates a new function that will force the this inside the function to be the parameter passed to bind().

Here's an example that shows how to use bind to pass a member method around that has the correct this:

var myButton = {
  content: 'OK',
  click() {
    console.log(this.content + ' clicked');
  }
};

myButton.click();

var looseClick = myButton.click;
looseClick(); // not bound, 'this' is not myButton - it is the globalThis

var boundClick = myButton.click.bind(myButton);
boundClick(); // bound, 'this' is myButton

Which prints out:

OK clicked
undefined clicked
OK clicked

You can also add extra parameters after the 1st (this) parameter and bind will pass in those values to the original function. Any additional parameters you later pass to the bound function will be passed in after the bound parameters:

// Example showing binding some parameters
var sum = function(a, b) {
  return a + b;
};

var add5 = sum.bind(null, 5);
console.log(add5(10));

Which prints out:

15

Check out JavaScript Function bind for more info and interactive examples.

Update: ECMAScript 2015 adds support for => functions. => functions are more compact and do not change the this pointer from their defining scope, so you may not need to use bind() as often. For example, if you wanted a function on Button from the first example to hook up the click callback to a DOM event, the following are all valid ways of doing that:

var myButton = {
  ... // As above
  hookEvent(element) {
    // Use bind() to ensure 'this' is the 'this' inside click()
    element.addEventListener('click', this.click.bind(this));
  }
};

Or:

var myButton = {
  ... // As above
  hookEvent(element) {
    // Use a new variable for 'this' since 'this' inside the function
    // will not be the 'this' inside hookEvent()
    var me = this;
    element.addEventListener('click', function() { me.click() });
  }
};    

Or:

var myButton = {
  ... // As above
  hookEvent(element) {
    // => functions do not change 'this', so you can use it directly
    element.addEventListener('click', () => this.click());
  }
};
Tuesday, June 1, 2021
 
Laimoncijus
answered 7 Months ago
99

It is an "internal property" of the object. From ECMAScript 8.6.2:

This specification uses various internal properties to define the semantics of object values. These internal properties are not part of the ECMAScript language. They are defined by this specification purely for expository purposes. An implementation of ECMAScript must behave as if it produced and operated upon internal properties in the manner described here. The names of internal properties are enclosed in double square brackets [[ ]].

The statement, "These internal properties are not part of the ECMAScript language," means that internal properties are not identifiers that can be used in actual code -- internal properties are not accessible as members of the objects that contain them. However, they may be made accessible by particular functions or properties (e.g., some browsers are kind enough let you set and get [[Prototype]] through the __proto__ property, and the ES5 spec allows read-only access through Object.getPrototypeOf).

The use of double brackets over single brackets is probably to avoid any possible confusion with actual bracket notation (i.e., property access).

Thursday, June 10, 2021
 
Dail
answered 7 Months ago
15

constructor is a regular property of the prototype object (with the DontEnum flag set so it doesn't show up in for..in loops). If you replace the prototype object, the constructor property will be replaced as well - see this explanation for further details.

You can work around the issue by manually setting Obj2.prototype.constructor = Obj2, but this way, the DontEnum flag won't be set.

Because of these issues, it isn't a good idea to rely on constructor for type checking: use instanceof or isPrototypeOf() instead.


Andrey Fedorov raised the question why new doesn't assign the constructor property to the instance object instead. I guess the reason for this is along the following lines:

All objects created from the same constructor function share the constructor property, and shared properties reside in the prototype.

The real problem is that JavaScript has no built-in support for inheritance hierarchies. There are several ways around the issue (yours is one of these), another one more 'in the spirit' of JavaScript would be the following:

function addOwnProperties(obj /*, ...*/) {
    for(var i = 1; i < arguments.length; ++i) {
        var current = arguments[i];

        for(var prop in current) {
            if(current.hasOwnProperty(prop))
                obj[prop] = current[prop];
        }
    }
}

function Obj1(arg1) {
    this.prop1 = arg1 || 1;
}

Obj1.prototype.method1 = function() {};

function Obj2(arg1, arg2) {
    Obj1.call(this, arg1);
    this.test2 = arg2 || 2;
}

addOwnProperties(Obj2.prototype, Obj1.prototype);

Obj2.prototype.method2 = function() {};

This makes multiple-inheritance trivial as well.

Monday, July 12, 2021
 
SheppardDigital
answered 5 Months ago
100

Lets examine your code a little bit.

function Food(){}
function Bread(){}
function Sushi(){}
var basicFood = new Food();
Bread.prototype = basicFood;
Sushi.prototype = basicFood;

Note: When you set the same object as the prototype of two objects, augmentation in one prototype, will reflect in the other prototype as well. For example,

Bread.prototype = basicFood;
Sushi.prototype = basicFood;
Bread.prototype.testFunction = function() {
    return true;
}
console.log(Sushi.prototype.testFunction()); // true

Lets get back to your questions.

var bread = reconstructify(new Bread(), Bread);
var sushi = reconstructify(new Sushi(), Sushi);
console.log(sushi instanceof Bread);    // Why true?
console.log(bread instanceof Sushi);    // Why true?

As per the instanceof docs from MDN,

The instanceof operator tests whether an object has in its prototype chain the prototype property of a constructor.

So when we do something like

object1 instanceof object2

JavaScript will try to find if the prototype of the object2 is in the prototype chain of object1.

In this case, it will return true only when the Bread.prototype is in the prototype chain of sushi. We know that sushi is constructed from Sushi. So, it will take Sushi's prototype and check if it is equal to Bread's prototype. Since, they both point to the same basicFood object, that returns true. Same case for, bread instanceof Sushi as well.

So, the right way to inherit would be, like this

function Food()  {}
function Bread() {}
function Sushi() {}

Bread.prototype = Object.create(Food.prototype);
Bread.prototype.constructor = Bread;
Sushi.prototype = Object.create(Food.prototype);
Sushi.prototype.constructor = Sushi;

var bread = new Bread();
var sushi = new Sushi();

console.log(sushi instanceof Bread);  // false
console.log(bread instanceof Sushi);  // false
console.log(sushi.constructor);       // [Function: Sushi]
console.log(bread.constructor);       // [Function: Bread]
console.log(sushi instanceof Food);   // true
console.log(bread instanceof Food);   // true
console.log(sushi instanceof Sushi);  // true
console.log(bread instanceof Bread);  // true
Monday, August 9, 2021
 
Thiago Custodio
answered 4 Months ago
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