W3cubDocs

/JavaScript

super

The super keyword is used to access properties on an object literal or class's [[Prototype]], or invoke a superclass's constructor.

The super.prop and super[expr] expressions are valid in any method definition in both classes and object literals. The super(...args) expression is valid in class constructors.

Syntax

super([arguments]) // calls the parent constructor.
super.propertyOnParent
super[expression]

Description

The super keyword can be used in two ways: as a "function call" (super(...args)), or as a "property lookup" (super.prop and super[expr]).

Note: super is a keyword and these are special syntactic constructs. super is not a variable that points to the prototype object. Attempting to read super itself is a SyntaxError.

const child = {
  myParent() {
    console.log(super); // SyntaxError: 'super' keyword unexpected here
  },
};

In the constructor body of a derived class (with extends), the super keyword may appear as a "function call" (super(...args)), which must be called before the this keyword is used, and before the constructor returns. It calls the parent class's constructor and binds the parent class's public fields, after which the derived class's constructor can further access and modify this.

The "property lookup" form can be used to access methods and properties of an object literal's or class's [[Prototype]]. Within a class's body, the reference of super can be either the superclass's constructor itself, or the constructor's prototype, depending on whether the execution context is instance creation or class initialization. See the Examples section for more details.

Note that the reference of super is determined by the class or object literal super was declared in, not the object the method is called on. Therefore, unbinding or re-binding a method doesn't change the reference of super in it (although they do change the reference of this). You can see super as a variable in the class or object literal scope, which the methods create a closure over. (But also beware that it's not actually not a variable, as explained above.)

When setting properties through super, the property is set on this instead.

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'

Accessing super in class field declaration

super can also be accessed during class field initialization. The reference of super depends on whether the current field is an instance field or a static field.

class Base {
  static baseStaticField = 90;
  baseMethod() {
    return 10;
  }
}

class Extended extends Base {
  extendedField = super.baseMethod(); // 10
  static extendedStaticField = super.baseStaticField; // 90
}

Note that instance fields are set on the instance instead of the constructor's prototype, so you can't use super to access the instance field of a superclass.

class Base {
  baseField = 10;
}

class Extended extends Base {
  extendedField = super.baseField; // undefined
}

Here, extendedField is undefined instead of 10, because baseField is defined as an own property of the Base instance, instead of Base.prototype. super, in this context, only looks up properties on Base.prototype, because that's the [[Prototype]] of Extended.prototype.

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'.

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.

const obj1 = {
  method1() {
    console.log('method 1');
  }
}

const obj2 = {
  method2() {
    super.method1();
  }
}

Object.setPrototypeOf(obj2, obj1);
obj2.method2(); // logs "method 1"

Methods that read super.prop do not behave differently when bound to other objects

Accessing super.x behaves like Reflect.get(Object.getPrototypeOf(objectLiteral), "x", this), which means the property is always seeked on the object literal/class declaration's prototype, and unbinding and re-binding a method won't change the reference of super.

class Base {
  baseGetX() {
    return 1;
  }
}
class Extended extends Base {
  getX() {
    return super.baseGetX();
  }
}

const e = new Extended();
console.log(e.getX()); // 1
const { getX } = e;
console.log(getX()); // 1

The same happens in object literals.

const parent1 = { prop: 1 };
const parent2 = { prop: 2 };

const child = {
  myParent() {
    console.log(super.prop);
  },
};

Object.setPrototypeOf(child, parent1);
child.myParent(); // logs "1"

const myParent = child.myParent;
myParent(); // still logs "1"

const anotherChild = { __proto__: parent2, myParent };
anotherChild.myParent(); // still logs "1"

Only resetting the entire inheritance chain will change the reference of super.

class Base {
  baseGetX() { return 1; }
  static staticBaseGetX() { return 3; }
}
class AnotherBase {
  baseGetX() { return 2; }
  static staticBaseGetX() { return 4; }
}
class Extended extends Base {
  getX() { return super.baseGetX(); }
  static staticGetX() { return super.staticBaseGetX(); }
}

const e = new Extended();
// Reset instance inheritance
Object.setPrototypeOf(Extended.prototype, AnotherBase.prototype);
console.log(e.getX()); // Logs "2" instead of "1", because the prototype chain has changed
console.log(Extended.staticGetX()); // Still logs "3", because we haven't modified the static part yet
// Reset static inheritance
Object.setPrototypeOf(Extended, AnotherBase);
console.log(Extended.staticGetX()); // Now logs "4"

Setting super.prop will set the property on this instead

Setting properties of super, such as super.x = 1, behaves like Reflect.set(Object.getPrototypeOf(objectLiteral), "x", 1, this). This is one of the cases where understanding super as simply "reference of the prototype object" falls short, because it actually sets the property on this instead.

class A {}
class B extends A {
  setX() {
    super.x = 1;
  }
}

const b = new B();
b.setX();
console.log(b); // B { x: 1 }
console.log(Object.hasOwn(b, "x")); // true

super.x = 1 will look for the property descriptor of x on A.prototype (and invoke the setters defined there), but the this value will be set to this, which is b in this context. You can read Reflect.set for more details on the case when target and receiver differ.

This means that while methods that get super.prop are usually not susceptible to changes in the this context, those that set super.prop are.

/* Reusing same declarations as above */

const b2 = new B();
b2.setX.call(null); // TypeError: Cannot assign to read only property 'x' of object 'null'

However, super.x = 1 still consults the property descriptor of the prototype object, which means you cannot rewrite non-writable properties, and setters will be invoked.

class X {
  constructor() {
    // Create a non-writable property
    Object.defineProperty(this, 'prop', {
      configurable: true,
      writable: false,
      value: 1,
    });
  }
}

class Y extends X {
  constructor() {
    super();
  }
  foo() {
    super.prop = 2;   // Cannot overwrite the value.
  }
}

const y = new Y();
y.foo(); // TypeError: "prop" is read-only
console.log(y.prop); // 1

Specifications

Browser compatibility

Desktop Mobile Server
Chrome Edge Firefox Internet Explorer Opera Safari WebView Android Chrome Android Firefox for Android Opera Android Safari on IOS Samsung Internet Deno Node.js
super
42
13
45
No
29
7
42
42
45
29
7
4.0
1.0
6.0.0

See also

© 2005–2022 MDN contributors.
Licensed under the Creative Commons Attribution-ShareAlike License v2.5 or later.
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/super