Class: super in method

Inside a class use super to access parent methods

use of super without extends fails (already when transpiling)

class A {hasSuper() { return super; }} assert.equal(new A().hasSuper(), false);

super with extends calls the method of the given name of the parent class

class A {hasSuper() { return true; }} class B extends A {hasSuper() { return super.hasSuper; }} assert.equal(new B().hasSuper(), true);

when overridden a method does NOT automatically call its super method

class A {hasSuper() { return true; }} class B extends A {hasSuper() { return 'nothing'; }} assert.equal(new B().hasSuper(), void 0);

super works across any number of levels of inheritance

class A {iAmSuper() { return true; }} class B extends A {} class C extends B {iAmSuper() { return iAmSuper(); }} assert.equal(new C().iAmSuper(), true);

accessing an undefined member of the parent class returns undefined

class A {} class B extends A {getMethod() { return super.constructor; }} assert.equal(new B().getMethod(), void 0);

Related Katas

Class

• creation
• accessors
• static
• extends
• more extends
• super in method
• super in constructor

Difficulty Level

INTERMEDIATE

First Published

21 April 2015

Stats

5 tests to solve