Javascript ES6 Iterator protocol and forEach

Question

Currently I am sharpening my ES6 skills a bit. I am looking into Iterator/Generator-syntax. I have a working example of

class Library {
    constructor(){
        this._books = [];
    }
    addBook(book){
        this._books.push(book);
    }
    get books() {
        return this._books;
    }
    *[Symbol.iterator]() {
        for(let i=0; i<this._books.length; i++) {
            yield this._books[i];
        }
    }
}

l = new Library();
l.addBook("Book1");
l.addBook("Book2");

for(let book of l){
    console.log(book);
}

Where everything works fine. But my first approach was trying something like

*[Symbol.iterator]() {
    this._books.forEach(
        book => yield book
    )
}

Which is (obviously) not correct. Is there besides looping with for or while a more concise way, to write this?

You have to use yield in a generator function, not in a non-generator function, which is what you are doing in the forEach callback. — trincot
– trincot, Commented Nov 10, 2017 at 21:27
Your yield in the non-working code is inside the callback to .forEach. That callback function is not a generator. — Pointy
– Pointy, Commented Nov 10, 2017 at 21:28
@Will How is WAG related? You really should take care of your ass ...? — Teemu
– Teemu, Commented Nov 10, 2017 at 21:28
It seems return _books[Symbol.iterator]() should work, but it doesn't. Unsure why. Devil in the implementation, I guess. — user1228
– user1228, Commented Nov 10, 2017 at 21:38

trincot · Accepted Answer · 2017-11-10 21:40:51Z

3

I would do it using yield*:

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

answered Nov 10, 2017 at 21:40

trincot

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2 Comments

Kyle Richardson Over a year ago

I learn something almost every time I see one of your post. Keep up the good work!

Thomas Junk Over a year ago

Whoha. Nice! That's the sort of code, I was looking for.

Kyle Richardson · Accepted Answer · 2017-11-10 21:45:23Z

2

~~I believe the following would be the most idiomatic approach.~~ Please refer to trincot's answer.

*[Symbol.iterator]() {
    for(let book of this._books) {
        yield book;
    }
}

edited Nov 10, 2017 at 21:45

answered Nov 10, 2017 at 21:36

Kyle Richardson

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2 Comments

Thomas Junk Over a year ago

Okay! Seems legit. at least nicer, than what I first tried.

Kyle Richardson Over a year ago

trincot's answer is better.

loganfsmyth · Accepted Answer · 2017-11-10 23:30:39Z

1

If you're literally passing through the iterator, you can also do

[Symbol.iterator]() {
  return this._books[Symbol.iterator]();
}

and skip needing the generator in the first place.

answered Nov 10, 2017 at 23:30

loganfsmyth

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1 Comment

Thomas Junk Over a year ago

Yes, you are right. But the point was more of playing with syntax and expressiveness of the language to get a feeling for what works in which way. I saw the Generator example and began toying around. Not that the code makes any bigger "sense" ;-)

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