constructor
The constructor method is a special method of a class for creating and initializing an object instance of that class.
Note: This page introduces the constructor syntax. For the constructor property present on all objects, see Object.prototype.constructor .
Try it
Syntax
constructor() /* … */ > constructor(argument0) /* … */ > constructor(argument0, argument1) /* … */ > constructor(argument0, argument1, /* … ,*/ argumentN) /* … */ >
There are some additional syntax restrictions:
- A class method called constructor cannot be a getter, setter, async, or generator.
- A class cannot have more than one constructor method.
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(); // Hello, my name is Otto
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); >
Note: The difference between an explicit constructor like the one above and the default constructor is that the latter doesn’t actually invoke the array iterator through argument spreading.
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; > >
Using new on a class goes through the following steps:
- (If it’s a derived class) The constructor body before the super() call is evaluated. This part should not access this because it’s not yet initialized.
- (If it’s a derived class) The super() call is evaluated, which initializes the parent class through the same process.
- The current class’s fields are initialized.
- The constructor body after the super() call (or the entire body, if it’s a base class) is evaluated.
Within the constructor body, you can access the object being created through this and access the class that is called with new through new.target . Note that methods (including getters and setters) and the prototype chain are already initialized on this before the constructor is executed, so you can even access methods of the subclass from the constructor of the superclass. However, if those methods use this , the this will not have been fully initialized yet. This means reading public fields of the derived class will result in undefined , while reading private fields will result in a TypeError .
new (class C extends class B constructor() console.log(this.foo()); > > #a = 1; foo() return this.#a; // TypeError: Cannot read private member #a from an object whose class did not declare it // It's not really because the class didn't declare it, // but because the private field isn't initialized yet // when the superclass constructor is running > >)();
The constructor method may have a return value. While the base class may return anything from its constructor, the derived class must return an object or undefined , or a TypeError will be thrown.
class ParentClass constructor() return 1; > > console.log(new ParentClass()); // ParentClass <> // The return value is ignored because it's not an object // This is consistent with function constructors class ChildClass extends ParentClass constructor() return 1; > > console.log(new ChildClass()); // TypeError: Derived constructors may only return object or undefined
If the parent class constructor returns an object, that object will be used as the this value on which class fields of the derived class will be defined. This trick is called «return overriding», which allows a derived class’s fields (including private ones) to be defined on unrelated objects.
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
Async methods, generator methods, accessors, and class fields are forbidden from being called constructor . Private names cannot be called #constructor . Any member named constructor must be a plain method.
Examples
Using the constructor
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; > >
Calling super in a constructor bound to a different prototype
super() calls the constructor that’s the prototype of the current class. If you change the prototype of the current class itself, super() will call the constructor that’s the new prototype. Changing the prototype of the current class’s prototype property doesn’t affect which constructor super() calls.
class Polygon constructor() this.name = "Polygon"; > > class Rectangle constructor() this.name = "Rectangle"; > > class Square extends Polygon constructor() super(); > > // Make Square extend Rectangle (which is a base class) instead of Polygon Object.setPrototypeOf(Square, Rectangle); const newInstance = new Square(); // newInstance is still an instance of Polygon, because we didn't // change the prototype of Square.prototype, so the prototype chain // of newInstance is still // newInstance --> Square.prototype --> Polygon.prototype console.log(newInstance instanceof Polygon); // true console.log(newInstance instanceof Rectangle); // false // However, because super() calls Rectangle as constructor, the name property // of newInstance is initialized with the logic in Rectangle console.log(newInstance.name); // Rectangle
Specifications
Browser compatibility
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