ES6 Destructuring — refer to this/self

Asked Sep 28 '18 at 07:34

Active Sep 28 '18 at 12:16

Viewed 159 times

Got a class like this:

class Item {
  constructor (next = null) {
    this.next = next
  }

  get self () { return this; }

  *[Symbol.iterator] () {
    yield this;
    if (this.next) yield *this.next;
  }
}

What is basically a linked list. Lets assume the list has at least three items and that root holds a reference to the head.

Right now, I can iterate like that and have a reference to both, the current item itself and its successor.

for (let { self, next } of root) {

}

Can the above be done without having the »self getter« withing the class, with destructuring only? So I have tried for (let {this, next} of root) but that does not work.

So can I have a reference to the »source object« with destructuring ?

edited Sep 28 '18 at 12:16

asked Sep 28 '18 at 07:34

philipp

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Does it work without the destructuring? – evolutionxbox Sep 28 '18 at 07:36
yes. but that I need to access the successor like `self.next` all the time. – philipp Sep 28 '18 at 07:42
1

An object doesn't have a property that references itself, so there is nothing to destructure. All you could do is split this up into two parts: `for (let self of root) { const {next} = self; }`. – Felix Kling Sep 28 '18 at 07:59
1

Kind of a duplicate of https://stackoverflow.com/q/29051011/218196 maybe. – Felix Kling Sep 28 '18 at 08:00
@FelixKling Already got that… – philipp Sep 28 '18 at 08:02
I don't think you can, this might be a limitation of destructuring. – evolutionxbox Sep 28 '18 at 08:48
That is why I am asking, so this might be the answer ;) – philipp Sep 28 '18 at 08:50
Btw, you don't need to call the `[Symbol.iterator]()`, you can just do `if (this.next) yield* this.next;` – Bergi Sep 28 '18 at 09:37
@Bergi Thanks for this one, much better so… ;) – philipp Sep 28 '18 at 12:16

ES6 Destructuring — refer to this/self

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