In ES6, what is the behavior of using super in a method when extending a class?
- Calls the superclass constructor
- References the superclass prototype
- Creates a new instance of the superclass
- Invokes the superclass method
In ES6, when using super in a method of a subclass, it refers to the superclass, allowing access to its methods and properties. It is often used to call the superclass constructor using super() or invoke methods on the superclass. This facilitates extending the functionality of the superclass.
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