Calling the Parent with super in TypeScript in TypeScript

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(): void {
        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(): void, but the body is empty. Inside it:

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:

• Dog's constructor declares protected name: string, so Puppy may read this.name but code outside the classes may not.
• Puppy has no constructor of its own, so it inherits Dog's: new Puppy("Buddy") sets the name.

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

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

const rex = new Puppy("Rex");
rex.bark();
const 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