Generator return doesn't work in for-await-of loop

Question

Generator with return statement skips the execution of for-await-of loop.

(async () => {
  const gen = async function* () {
     return { newState: "FAILURE" };
  };
  for await (const { newState } of gen()) {
    console.log("Yey! new state is:", newState);
    // do other stuff
  }
})();

In the case written above, the whole for-await-of construction will never console.log anything. But when you swap return for yield (yield { newState: 'FAILURE' }) everything work as intended.

(async () => {
  const gen = async function* () {
    yield { newState: "FAILURE" };
  };
  for await (const { newState } of gen()) {
    console.log("Yey! new state is:", newState);
    // do other stuff
  }
})();

WHY?

Bergi · Accepted Answer · 2021-06-09 12:57:00Z

1

The iterator never yielded anything before completing, so the loop didn't have anything to iterate - the sequence was empty. Notice you can yield multiple times but return only once.

A for … of loop ignores the return value of an iterator. It's only useful when you are manually advancing the iterator, or when using yield*.

async function* generate() {
  return { newState: "FAILURE" };
};
const generator = (async function*() {
  const { newState } = yield* generate()
  console.log("Yey! new state is:", newState);
  return "some value";
})();
generator.next().then(console.log);

This is no different for asynchronous iterators than for synchronous iterators.

edited Jun 9, 2021 at 12:57

answered Jun 9, 2021 at 12:49

Bergi

671k162 gold badges1k silver badges1.5k bronze badges

Sign up to request clarification or add additional context in comments.

9 Comments

Src Over a year ago

Can you please provide some docs where explicitly said that/why the for-await-of ignores return?

Bergi Over a year ago

@Src It doesn't ignore the return statement - it stops as expected. It just doesn't have anywhere to put the return value - a for loop is not an expression with a result value.

Src Over a year ago

According to the docs, yield and return both returns values, but return instead of yield returns {done:true}. So why it isn’t act like yield in this case?

Src Over a year ago

Also how loop knows when to stop, because if we swap return with yield, we still get the correct result, taking into account that we don’t manually said to stop iterating, which you normally do with return

Bergi Over a year ago

@FaizanurRahman Simply because it's sometimes useful, mostly when using generators as coroutines that have a result, not when using them to iterate something.

|

Collectives™ on Stack Overflow

Generator return doesn't work in for-await-of loop

1 Answer 1

9 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

1 Answer 1

9 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related