super()

devfish·2023년 1월 16일

Javascript

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What is super?

We call super() inside of a subclass' constructor method in order to call the parent constructor... and that's pretty much what super is in JavaScript: a reference to the parent prototype

we use super to reference the parent prototype's constructor function

class Resource { constructor() { } }

class Widget extends Resource {
    constructor() {
        super();
        if (super.constructor === Resource.prototype.constructor) {
            console.log("It's true!");
        }
    }
}

must be specified before using this
if you have a constructor method in the sub-class, you need one in the parent class
- JS only attaches an object instance to this once you get to the highest prototype in the chain: Object
- Default return value from constructor method is this, and if there is no parent constructor, you will have no this value and it will stay undefined
JS will create a no-op version of a constructor for you if you don't explicitly create one in the lower level constructor (in subclass):
```
constructor(...args) {
   super(...args);
}
```

Auto-context switching!

JS will switch the current context for this for you, even if it's in the parent class!

class Parent {
    constructor() { }
    whoami() {
        console.log(this);  // will print out the current SUBCLASS object
    }
}

class Child extends Parent {
    constructor() { super(); }
    whoami() {
        console.log(this);  // will print out the current Child object
        super.whoami();
    }
}

let w = new Child();
w.whoami();

So you can do this instead:

class Bee {
    constructor() { }
    whoami() {
        console.log(this);  // will print out the current SUBCLASS object
    }
}

class WorkerBee extends Bee {
   // constructor() { super(); }
}

let w = new WorkerBee(); 
console.log(w.whoami());