The constructor of a parent class will always be invoked before the constructor of its child

The constructor method is a special method of a class for creating and initializing an object instance of that class.

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Syntax

constructor() { /* … */ } constructor(argument0) { /* … */ } constructor(argument0, argument1) { /* … */ } constructor(argument0, argument1, /* … ,*/ argumentN) { /* … */ }

Description

A constructor enables you to provide any custom initialization that must be done before any other methods can be called on an instantiated object.

class Person { constructor(name) { this.name = name; } introduce() { console.log(`Hello, my name is ${this.name}`); } } const otto = new Person('Otto'); otto.introduce();

If you don't provide your own constructor, then a default constructor will be supplied for you. If your class is a base class, the default constructor is empty:

If your class is a derived class, the default constructor calls the parent constructor, passing along any arguments that were provided:

constructor(...args) { super(...args); }

That enables code like this to work:

class ValidationError extends Error { printCustomerMessage() { return `Validation failed :-( (details: ${this.message})`; } } try { throw new ValidationError("Not a valid phone number"); } catch (error) { if (error instanceof ValidationError) { console.log(error.name); // This is Error instead of ValidationError! console.log(error.printCustomerMessage()); } else { console.log('Unknown error', error); throw error; } }

The ValidationError class doesn't need an explicit constructor, because it doesn't need to do any custom initialization. The default constructor then takes care of initializing the parent Error from the argument it is given.

However, if you provide your own constructor, and your class derives from some parent class, then you must explicitly call the parent class constructor using super. For example:

class ValidationError extends Error { constructor(message) { super(message); // call parent class constructor this.name = 'ValidationError'; this.code = '42'; } printCustomerMessage() { return `Validation failed :-( (details: ${this.message}, code: ${this.code})`; } } try { throw new ValidationError("Not a valid phone number"); } catch (error) { if (error instanceof ValidationError) { console.log(error.name); // Now this is ValidationError! console.log(error.printCustomerMessage()); } else { console.log('Unknown error', error); throw error; } }

There can be only one special method with the name constructor in a class. Having more than one occurrence of a constructor method in a class will throw a SyntaxError error. Having a getter or setter called constructor is also a SyntaxError.

The constructor follows normal method syntax, so parameter default values, rest parameters, etc. can all be used.

class Person { constructor(name = 'Anonymous') { this.name = name; } introduce() { console.log(`Hello, my name is ${this.name}`); } } const person = new Person(); person.introduce(); // Hello, my name is Anonymous

The constructor must be a literal name. Computed properties cannot become constructors.

class Foo { // This is a computed property. It will not be picked up as a constructor. ['constructor']() { console.log('called'); this.a = 1; } } const foo = new Foo(); // No log console.log(foo); // Foo {} foo.constructor(); // Logs "called" console.log(foo); // Foo { a: 1 }

Examples

Using the constructor method

This code snippet is taken from the classes sample (live demo).

class Square extends Polygon { constructor(length) { // Here, it calls the parent class' constructor with lengths // provided for the Polygon's width and height super(length, length); // NOTE: In derived classes, `super()` must be called before you // can use `this`. Leaving this out will cause a ReferenceError. this.name = 'Square'; } get area() { return this.height * this.width; } set area(value) { this.height = value ** 0.5; this.width = value ** 0.5; } }

Another example

Here the prototype of Square class is changed—but the constructor of its base class Polygon is still called when a new instance of a square is created.

class Polygon { constructor() { this.name = "Polygon"; } } class Square extends Polygon { constructor() { super(); } } class Rectangle {} Object.setPrototypeOf(Square.prototype, Rectangle.prototype); console.log(Object.getPrototypeOf(Square.prototype) === Polygon.prototype); //false console.log(Object.getPrototypeOf(Square.prototype) === Rectangle.prototype); //true const newInstance = new Square(); console.log(newInstance.name); //Polygon

Specifications

Specification
ECMAScript Language Specification
# sec-static-semantics-constructormethod

Browser compatibility

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See also

Does a child constructor always invoke a parent constructor?

If the child class constructor does not call super , the parent's constructor with no arguments will be implicitly called. If parent class implements a constructor with arguments and has no a constructor with no arguments, then the child constructors must explicitly call a parents constructor.

Is parent or child constructor called first?

While object creation of a class, the default constructor of that class is automatically called to initialize the members of the class. In case of inheritance if we create an object of child class then the parent class constructor will be called before child class constructor.

Which constructor executes first the parent constructor or the child constructor?

If we create an object of the child class in the program, then, the body of the constructor of the parent class will execute first, then the body of the child class will be executed. In simple words, we can say that the parent constructor gets called first, then of the child class.

Can be used to invoke the parent class constructor?

super() can be used to invoke immediate parent class constructor.