Inheritance and super

In short: class Child extends Parent makes Child inherit every method Parent has — an instance of Child is also, genuinely, an instance of Parent. Inside Child's constructor, super(...) calls Parent's constructor, and it must run before this is used — a subclass constructor has no usable this until its parent's part of the object has been built. Inside a method, super.method() calls the parent's version of that method specifically, even when the child has overridden it with its own.

Extend a class and call its parent's method

super(title) builds the parent portion of the object; super.describe() later calls Song's own describe, and the child's version extends that result:

s.describe() prints "Song: Blue Hour (live at The Grand)" — LiveSong's describe calls super.describe() to reuse Song's formatting, then adds its own venue text. s instanceof Song is true: an instance of a subclass is also an instance of every class it extends, all the way up.

The trap: using this before calling super()

In a subclass constructor, this isn't ready until super() has actually run — reading or writing it any earlier throws immediately:

This throws ReferenceError: Must call super constructor in derived class before accessing 'this' or returning from derived constructor. The fix: super(title) must be the first statement in a subclass constructor, before any use of this.

Quick practice

🎯 Your turn

Given class Song { constructor(title) { this.title = title; } } class LiveSong extends Song { constructor(title, venue) { this.venue = venue; super(title); } }, explain exactly why new LiveSong("a", "b") throws an error, and how to fix the constructor.

Write your explanation and submit it — there's no code to run here, just your reasoning.