javascript - ES6 What does super() actually do in constructor function? - Stack Overflow

admin管理员组

文章数量:1340984

! Hola, amigos. I have this little class inheritance structure

class Point {
    constructor(x, y) {
        this.x = x;
        this.y = y;
    }
    toString() {
        return '(' + this.x + ', ' + this.y + ')';
    }
}

class ColorPoint extends Point {
    constructor(x, y, color) {
        super(x, y); 
        this.color = color;
    }
    toString() {
        return super.toString() + ' in ' + this.color; 
    }
}

let newObj = new ColorPoint(25, 8, 'green');

It piles to this jsfiddle

I get how it works in es6 in a silly way. But could somebody explain how does it work under the hood in es5. In a simpler form.

! Hola, amigos. I have this little class inheritance structure

class Point {
    constructor(x, y) {
        this.x = x;
        this.y = y;
    }
    toString() {
        return '(' + this.x + ', ' + this.y + ')';
    }
}

class ColorPoint extends Point {
    constructor(x, y, color) {
        super(x, y); 
        this.color = color;
    }
    toString() {
        return super.toString() + ' in ' + this.color; 
    }
}

let newObj = new ColorPoint(25, 8, 'green');

It piles to this jsfiddle

I get how it works in es6 in a silly way. But could somebody explain how does it work under the hood in es5. In a simpler form.

• javascript
• class
• constructor
• ecmascript-6
• super

Share Improve this question Follow asked Mar 3, 2017 at 15:00 Sergio NikolaevSergio Nikolaev 78988 silver badges1616 bronze badges 3

• 4 It invokes constructor() on the parent class. At the simple level, there's really only so much to say. Is there something in particular you're curious about? – S McCrohan Commented Mar 3, 2017 at 15:06
• 1 @SMcCrohan, maybe try to explain not in a simpler form. It invokes constructor() on the parent class - that was already grasped. But how does it invokes it? How does these arguments get passed from subclass to superclass through super() . I understand these steps (maybe) but not the actual mechanics behind them. Why cant we just use .__proto__ and .call or .apply? – Sergio Nikolaev Commented Mar 3, 2017 at 15:30
• Can anyone explain what happens here stackoverflow./questions/54685556/… in context of super – Umair Abid Commented Feb 14, 2019 at 11:55

Add a ment  | 

1 Answer 1

Sorted by: Reset to default 15

super(…); is basically sugar for this = new ParentConstructor(…);. Where ParentConstructor is the extended class, and this = is the initialisation of the this keyword (well, given that that's forbidden syntax, there's a bit more than sugar to it). And actually it will inherit from the proper new.target.prototype instead of ParentConstructor.prototype like it would from new. So no, how it works under the hood does not pare to ES5 at all, this is really a new feature in ES6 classes (and finally enables us to properly subclass builtins).

本文标签： javascriptES6 What does super() actually do in constructor functionStack Overflow