Does super() map to __proto__ under the hood?

问题

I understand that classes in ES6 are really syntactic sugar. Is the super() call really just calling proto? (Is it mapped to the [[prototype]] object?)

回答1:

It's a bit more than that. It also remembers where the method was defined.

const example = {
    method() {
        return super.method();
    }
}

is syntactic sugar for

const example = {
    method() {
        return Object.getPrototypeOf(example).method.call(this);
    }
}

and

class Example {
    method() {
        return super.method();
    }
}

is syntactic sugar for

class Example {
    method() {
        return Object.getPrototypeOf(Example.prototype).method.call(this);
    }
}

As for super() calls in constructors, it similarly uses Object.getPrototypeOf on the constructor, but does a bit more there.

Is it mapped to the [[prototype]] object?

Yes. Not to the [[prototype]] of the object that the function was called on (this), but to the [[prototype]] of the object that the function was defined in.

来源：https://stackoverflow.com/questions/49822056/does-super-map-to-proto-under-the-hood