In ES6, what happens to the arguments in the first call to an iterator's `next` method?

Question

If you have an generator like,

function* f () {
  // Before stuff.
  let a = yield 1;
  let b = yield 2;
  return [a,b];
}

And, then run

Answer 1

From MDN Iterators and generators.

A value passed to next() will be treated as the result of the last yield expression that paused the generator.

Answers:

Does the argument in the first call to g.next get lost?

Since there is no last yield expression that paused the generator on the first call this value is essentially ignored. You can read more in the ECMAScript 2015 Language Specification.

What happens to them?

On subsequent calls of next() the value passed will be used as the return value of the last yield expression that paused the generator.

Using the above example, how do I set a?

You can do as LJHarb suggested.

"use strict";

let f = function*() {
	let a = yield 1;
	let b = yield 2;
	return [a, b];
};

let g = f();

document.querySelector("#log_1").innerHTML = JSON.stringify(g.next());
document.querySelector("#log_2").innerHTML = JSON.stringify(g.next(123));
document.querySelector("#log_3").innerHTML = JSON.stringify(g.next(456));

<div id="log_1"></div>
<div id="log_2"></div>
<div id="log_3"></div>

Answer 2

Values passed into the first 'next()' call are ignored. Look at the last test (line 34) on this ES6 TDD Coding Kata

For those confused on how a & b are getting set, it might be a good idea to look at the "Advanced Generators" section of Iterators & Generators

Answer 3

Try:

var g = f();
// this question is over this value.
g.next(); // returns: { value: 1, done: false }
g.next(123); // returns: { value: 2, done: false }
g.next(456); // returns: { value: [123, 456], done: true }