Q
How do you define a class in JavaScript ES6?

Answer & Solution

Answer: Option A
Solution:
Classes in ES6 are defined using the class keyword followed by the class name and a constructor method.
Related Questions on Average

How do you access the value of the name property in the object var person = {name: 'John', age: 30};?

A). person['name']

B). person.name

C). person->name

D). person::name

What is the output of console.log(person.greet()); if var person = {greet: function() {return 'Hello';}};?

A). Hello

B). Undefined

C). Error

D). Hello

What does Object.keys(person) return if var person = {name: 'John', age: 30};?

A). ['name', 'age']

B). ['John', 30]

C). {'name', 'age'}

D). {'John', 30}

How can you delete the age property from the person object?

A). delete person.age;

B). remove person.age;

C). person.age = null;

D). person.age.delete;

What is the correct way to create an object in JavaScript?

A). var obj = {};

B). var obj = [];

C). var obj = ();

D). var obj = <>;

How do you iterate over all properties of an object var person = {name: 'John', age: 30};?

A). for (var key in person)

B). for (var key of person)

C). forEach(key in person)

D). forEach(key of person)

How do you define a method named greet in an object var person = {};?

A). person.greet = function() {};

B). person.greet: function() {};

C). person.greet = () {};

D). person.greet = function {};

How do you create an instance of the Car object using the constructor function?

A). var myCar = new Car('Toyota', 'Corolla');

B). var myCar = Car('Toyota', 'Corolla');

C). var myCar = create Car('Toyota', 'Corolla');

D). var myCar = construct Car('Toyota', 'Corolla');

How do you add a new property gender with value male to the person object?

A). person.gender = 'male';

B). person.gender = male;

C). person.gender = ['male'];

D). person[gender] = 'male';

How do you call a method from a parent class in a subclass in JavaScript ES6?

A). super.methodName()

B). this.methodName()

C). parent.methodName()

D). base.methodName()