Answers for "javascript super"

7

super in javascirpt

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';
  }
}
Posted by: Guest on August-24-2020
1

extended class call method from super in javascript

// this is how we call method from a super class(i.e MAIN CLASS) to extended class
class Person {
 constructor(name, age) {
  this.name = name;
  this.age = age;
 }
greet() {
  console.log(`Hi, all my name is ${this.name}`);
 }
}

class Employee extends Person {
 constructor(name, age, employer) {
  super(name, age);  // NOTE : super must call before accessing (this)
  this.employer = employer;
 }

 greeting() {
  super.greet();  // this is how we call
  console.log(`I'm working at ${this.employer}`);
 }
}

let Steve = new Employee('Xman', 25 , 'Skynet');
Steve;
Steve.greeting(); // check by consoling MATE
Posted by: Guest on November-26-2020
1

javascript super

class Parent {
  constructor() {}
  method() {}
}
class Child extends Parent {
  constructor() {
    super() // Parent.constructor
    super.method() // Parent.method
  }
}
Posted by: Guest on December-25-2020
0

Super in javascript

super([arguments]); // calls the parent constructor.
super.functionOnParent([arguments]);
Posted by: Guest on April-19-2021

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