Fetch.run in batches without optimizations

[DenisNovac]

Hi everyone.

I have this piece of code:

def fetchRandomInt(max: Int) = Fetch(max, randomSource.source)

val fetchTrv: Fetch[IO, List[Int]] = List(10, 10, 10, 10, 10).traverse(fetchRandomInt)

println(Fetch.run(fetchTrv).unsafeRunSync)  // List(7, 7, 7, 7, 7)

Where randomSource is just a generator of random ints:

override def fetch(max: Int): IO[Option[Int]] =
  CF.delay {
    scala.util.Random.nextInt(max).some
  }

Is there any way to run this list of commands concurrently in one batch but without optimization by ID?

[sloshy]

Hi, obviously this is a late response and I apologize for that. I'm new to the project but I believe I can help you (or future people browsing/arriving from Google) here.

As you may have guessed from the other issue you posted (#306), the Fetch library uses a cache (DataCache) with every call to .run, with the default cache being a brand new InMemoryCache. The behavior of this cache is it effectively deduplicates requests within a single Fetch run. The implementation of .traverse here means that this cache is shared amongst each request so you see this behavior. If you do not traverse or flatMap them and instead run them individually (as .run without passing in any cache), you should see a random number generated each time, as you expect.

[DenisNovac]

@sloshy thank you for answer.