If you have an generator like,

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

And, then run

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

The first call to .next() to set a to 123 and the second call to set b to 456, however at the last call to .next() this is return,

{ value: [ 456, undefined ], done: true }

Does the argument in the first call to g.next get lost? What happens to them? Using the above example, how do I set a?

If you have an generator like,

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

And, then run

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

The first call to .next() to set a to 123 and the second call to set b to 456, however at the last call to .next() this is return,

{ value: [ 456, undefined ], done: true }

Does the argument in the first call to g.next get lost? What happens to them? Using the above example, how do I set a?

• javascript
• generator
• v8
• ecmascript-harmony

• Duplicate of stackoverflow./questions/20977379/… – Loamhoof Commented Jan 22, 2014 at 10:56
• Possible duplicate of Where argument of first next() call goes? – Bergi Commented Mar 30, 2016 at 17:32
• 1 In simple words ; when we pass argument to next(), it replaces previous yield expression return value with the one we provided.Because it is your first execution of function value 123 is simply discarded. – erhan355 Commented Nov 19, 2019 at 11:30

3 Answers 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 }

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

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>