When you have an async generator, you can call next(x) to return a result to the generator.

async function* generate() {
  for (let i = start; i <= 5; i++) {
    const result = yield i;
    console.log("result", result)
  }
}
;(async () => {
  const generator = generate()
  await generator.next(1)
  await generator.next(2)
})();

This will print 1 and 2 as this is the next parameter that is passed as a result back to the generator.

How to replicate this same behaviour but using await for ?

for await (let value of generate()) {
  // ... no way to send `next` parameter here
}