super
The super keyword is used to access and call functions on an object's parent.
The super.prop
and super[expr]
expressions are valid in any method definition in both classes and object literals.
Syntax
super([arguments]); // calls the parent constructor. super.functionOnParent([arguments]);
Description
When used in a constructor, the super
keyword appears alone and must be used before the this
keyword is used. The super
keyword can also be used to call functions on a parent object.
Examples
Using super
in classes
This code snippet is taken from the classes sample (live demo). Here super()
is called to avoid duplicating the constructor parts' that are common between Rectangle
and Square
.
class Rectangle { constructor(height, width) { this.name = 'Rectangle'; this.height = height; this.width = width; } sayName() { console.log('Hi, I am a ', this.name + '.'); } get area() { return this.height * this.width; } set area(value) { this._area = value; } } class Square extends Rectangle { constructor(length) { this.height; // ReferenceError, super needs to be called first! // Here, it calls the parent class's constructor with lengths // provided for the Rectangle'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 reference error. this.name = 'Square'; } }
Super-calling static methods
You are also able to call super on static methods.
class Rectangle { static logNbSides() { return 'I have 4 sides'; } } class Square extends Rectangle { static logDescription() { return super.logNbSides() + ' which are all equal'; } } Square.logDescription(); // 'I have 4 sides which are all equal'
Deleting super properties will throw an error
You cannot use the delete operator and super.prop
or super[expr]
to delete a parent class' property, it will throw a ReferenceError
.
class Base { foo() {} } class Derived extends Base { delete() { delete super.foo; // this is bad } } new Derived().delete(); // ReferenceError: invalid delete involving 'super'.
super.prop
cannot overwrite non-writable properties
When defining non-writable properties with e.g. Object.defineProperty
, super
cannot overwrite the value of the property.
class X { constructor() { Object.defineProperty(this, 'prop', { configurable: true, writable: false, value: 1 }); } } class Y extends X { constructor() { super(); } foo() { super.prop = 2; // Cannot overwrite the value. } } var y = new Y(); y.foo(); // TypeError: "prop" is read-only console.log(y.prop); // 1
Using super.prop
in object literals
Super can also be used in the object initializer / literal notation. In this example, two objects define a method. In the second object, super
calls the first object's method. This works with the help of Object.setPrototypeOf()
with which we are able to set the prototype of obj2
to obj1
, so that super
is able to find method1
on obj1
.
var obj1 = { method1() { console.log('method 1'); } } var obj2 = { method2() { super.method1(); } } Object.setPrototypeOf(obj2, obj1); obj2.method2(); // logs "method 1"
Specifications
Browser compatibility
Desktop | Mobile | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|
Chrome | Edge | Firefox | Internet Explorer | Opera | Safari | WebView Android | Chrome Android | Firefox for Android | Opera Android | Safari on IOS | Samsung Internet | |
super |
42 |
13 |
45 |
No |
29 |
7 |
42 |
42 |
45 |
29 |
7 |
4.0 |
See also
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Licensed under the Creative Commons Attribution-ShareAlike License v2.5 or later.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/super