Calculate the list of the futures and and return the result future

问题

I have a function which takes futures Future[A]* and I want it to return a Future[List[A]].

def singleFuture[T](futures: List[Future[A]]): Future[List[A]] = {
  val p = Promise[T]
  futures filter { _ onComplete { case x => p complete x /*????*/ } }
  p.future
}

And I also want the result future of type Future[List[A]] becomes completed immediately after the list futures List[Future[A]] have been completed.

That code doesn't work. I figure I should use flatMap here because there should be 2 internal loops: one for the future and one for promise. But how?

I'd like not use for comprehension here because I'd like to understand the process at the deeper lever.

回答1:

You can use foldRight to achieve this:

def singleFuture[A](futures: List[Future[A]]): Future[List[A]] = {

    val p = Promise[List[A]]()
    p.success(List.empty[A])

    val f = p.future // a future containing empty list.

    futures.foldRight(f) { 
       (fut, accum) =>  // foldRight means accumulator is on right.

        for {
           list <- accum;  // take List[A] out of Future[List[A]]
           a    <- fut     // take A out of Future[A]
        }
        yield (a :: list)   // A :: List[A]
    }
}

if any future in futures list fails, a <- fut will fail, resulting in accum being set to failed future.

If you want to avoid using for, you can either expand it to flatMap as follows:

  accum.flatMap( list => fut.map(a => a :: list))

Or you can use, async-await (noting that it is still an experimental feature).

def singleFuture[T](futures: List[Future[T]]): Future[List[T]] = async {

  var localFutures = futures
  val result = ListBuffer[T]()
  while (localFutures != Nil) {
        result += await { localFutures.head } 
        localFutures = localFutures.tail
  }
  result.toList
}

回答2:

This is already implemented for you:

def singleFuture[T](futures: List[Future[A]]): Future[List[A]] = Future.sequence(futures)

You can, of course, look at the implementation of sequence:

def sequence[A, M[_] <: TraversableOnce[_]](in: M[Future[A]])(implicit cbf: CanBuildFrom[M[Future[A]], A, M[A]], executor: ExecutionContext): Future[M[A]] = {
  in.foldLeft(Promise.successful(cbf(in)).future) {
    (fr, fa) => for (r <- fr; a <- fa.asInstanceOf[Future[A]]) yield (r += a)
  } map (_.result())
}

This can be simplified if you only want to deal with lists, and not with anything that has foldLeft:

def sequence[A](in: List[Future[A]]): Future[List[A]] = {
  in.foldRight[Future[List[A]](Promise.successful(Nil) {
    (fa, fr) => for { r <- fr; a <- fa } yield (a :: r)
  }
}

来源：https://stackoverflow.com/questions/20278133/calculate-the-list-of-the-futures-and-and-return-the-result-future

标签

scala

concurrency

future