How to run 2 for await loops in parallel?

Question

I have this code:

for await (const thing of signal()) {
    // logic
}

for await (const thing of anotherSignal()) {
    // logic
}

The problem is that these for awaits run forever. So the second one never executes. Is there a way to make these 2 loops run in parallel?

In parallel as in two threads or just running and reporting ASAP but leaving time for the other loop to also work? — VLAZ
– VLAZ, Commented Nov 9, 2021 at 8:49
@Zooly OP indicates those functions supply infinite promises. You cannot wait for all of them to finish. — VLAZ
– VLAZ, Commented Nov 9, 2021 at 8:50
If they never finish just put them both inside an async function and run, but you will want to also handle errors inside these functions. — Keith
– Keith, Commented Nov 9, 2021 at 8:54
Maybe it would be better to switch to a different abstraction, Observables in particular (e.g. rxjs.dev). — Felix Kling
– Felix Kling, Commented Nov 9, 2021 at 9:14

Bergi · Accepted Answer · 2021-11-21 22:52:55Z

1

Don't wait for first task to complete, Run them parallel, Like So:

let delay = ms => new Promise(ok => setTimeout(ok, ms));
async function* infinit_iter() {
  for (let i = 0;; i++) {
    await delay(3000)
    yield i
  }
}

async function run(name, iter) {
  for await (let i of iter)
    console.log(name, i)
}

run("signal:", infinit_iter())
run("anotherSignal:", infinit_iter())

edited Nov 21, 2021 at 22:52

Bergi

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answered Nov 10, 2021 at 6:25

Nur

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Bergi Over a year ago

What does "on demand" mean?

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