Calling the Parent with super in JavaScript in JavaScript

TL ; DR:

Inside a method of a class that extends another, super.bark() calls the parent's version of bark on the same object. This lets you extend the parent's behavior instead of replacing it.

class Puppy extends Dog {
    bark() {
        super.bark();   // do everything the parent did
        console.log("Yip! Said " + this.name);
    }
}

Why it matters

Overriding a method replaces the parent's behavior. Often you want to extend it instead: keep what the parent did and add something more. A typical example is a save method: the parent writes the row to the database, and a subclass runs the parent's save first and then writes a log line.

Your task

The Puppy class already overrides bark, but the body is empty. Inside Puppy's bark():

1. call the parent's version with super.bark();, so Woof! I am <name> still prints;
2. then console.log("Yip! Said " + this.name);.

Notes:

• super.bark() works inside a class method; super by itself refers to the parent class (Dog).
• this.name was set by Dog's constructor; Puppy has no constructor of its own, so it inherits it.

The driver (already written, it runs when you press Run):

let buddy = new Puppy("Buddy");
buddy.bark();

let rex = new Puppy("Rex");
rex.bark();
let dog = new Dog("Max");
dog.bark();

• buddy is a Puppy: its bark prints the parent's line, then its own.
• rex is another Puppy: the names come from the object, nothing is hard-coded.
• dog is a plain Dog: its bark is unchanged and prints only the parent's line, with no Yip!.

Expected output:

Woof! I am Buddy
Yip! Said Buddy
Woof! I am Rex
Yip! Said Rex
Woof! I am Max