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Why extending this Array does not need calling super?

I always thought that calling super() was practically like copy pasting the properties of the parent into the child class.

**Here super() is called to avoid duplicating the constructor parts' that are common between Rectangle and Square. ** MDN

I read on MDN that you need to extend a class and call super() to be able to. And also

The super keyword is used to access and call functions on an object's parent.

MDN

which I do not fully understand.

I tested those in this example:

class Logger extends Array {
    log() {
        let i = 0
        for (const el of this) {
            console.log(this[i], el)
            i++
        }
        return;
    }
}
a = new Logger(1, 2, 3)
a.log()
a.push(6)

But neither super() is needed for accessing this[i] or calling a.push() any hints folks?

about 4 years ago · Juan Pablo Isaza
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