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Scala 2.12 有 2Future.find种方法。

@deprecated("use the overloaded version of this method that takes a scala.collection.immutable.Iterable instead", "2.12.0")
def find[T](@deprecatedName('futurestravonce) futures: TraversableOnce[Future[T]])(@deprecatedName('predicate) p: T => Boolean)(implicit executor: ExecutionContext): Future[Option[T]]

及其重载版本

def find[T](futures: scala.collection.immutable.Iterable[Future[T]])(p: T => Boolean)(implicit executor: ExecutionContext): Future[Option[T]]

两者的描述相同

  /** Asynchronously and non-blockingly returns a `Future` that will hold the optional result
   *  of the first `Future` with a result that matches the predicate, failed `Future`s will be ignored.
   *
   * @tparam T        the type of the value in the future
   * @param futures   the `scala.collection.immutable.Iterable` of Futures to search
   * @param p         the predicate which indicates if it's a match
   * @return          the `Future` holding the optional result of the search
   */

所以我假设这些方法首先找到与给定列表中的Future参数匹配的完成p

但实际上只有第一个这样做。

  val start = System.currentTimeMillis
  val a = (1 to 3).reverse.iterator.map{ x =>
    Future{
      Thread.sleep(x * 10000)
      x
    }
  }
  val b = Future.find(a)(_.isInstanceOf[Int])
  b.foreach{ x =>
    println(x)
    println(System.currentTimeMillis - start) // 10020 
  }

该方法的弃用版本返回最快的版本。

  val a = (1 to 3).reverse.map{ x =>
    Future{
      Thread.sleep(x * 10000)
      x
    }
  }
  val b = Future.find(a)(_.isInstanceOf[Int])
  b.foreach{ x =>
    println(x)
    println(System.currentTimeMillis - start)
  }

重载版本返回最慢的版本。更准确地说，它只是从头到尾检查给定的列表，而不关心它们需要多长时间才能完成。

这是应该的吗？如果是这样，是否使用重复的或自己实施它只是关心他们完成时间的选择？

4

1 回答 1

1

你说得对，在2.12.x中期望 a的已弃用 Future.findTraversableOnce[Future[T]]的行为确实与 replace 不同Future.find。从下面粘贴的源代码中可以看出，前find一种方法利用PromisewithtryComplete有效地从输入集合中捕获第一个完整的未来，而后者采用简单的hasNext/next遍历：

@deprecated("use the overloaded version of this method that takes a scala.collection.immutable.Iterable instead", "2.12.0")
def find[T](@deprecatedName('futurestravonce) futures: TraversableOnce[Future[T]])(@deprecatedName('predicate) p: T => Boolean)(implicit executor: ExecutionContext): Future[Option[T]] = {
  val futuresBuffer = futures.toBuffer
  if (futuresBuffer.isEmpty) successful[Option[T]](None)
  else {
    val result = Promise[Option[T]]()
    val ref = new AtomicInteger(futuresBuffer.size)
    val search: Try[T] => Unit = v => try {
      v match {
        case Success(r) if p(r) => result tryComplete Success(Some(r))
        case _ =>
      }
    } finally {
      if (ref.decrementAndGet == 0) {
        result tryComplete Success(None)
      }
    }

    futuresBuffer.foreach(_ onComplete search)

    result.future
  }
}

def find[T](futures: scala.collection.immutable.Iterable[Future[T]])(p: T => Boolean)(implicit executor: ExecutionContext): Future[Option[T]] = {
  def searchNext(i: Iterator[Future[T]]): Future[Option[T]] =
    if (!i.hasNext) successful[Option[T]](None)
    else {
      i.next().transformWith {
        case Success(r) if p(r) => successful(Some(r))
        case other => searchNext(i)
      }
    }
  searchNext(futures.iterator)
}

实现自己的一种方法可能是将Future.firstCompletedOf带有添加谓词的方法扩展为如下所示：

def firstConditionallyCompletedOf[T](futures: List[Future[T]])(p: T => Boolean)(implicit ec: ExecutionContext): Future[T] = {
  val p = Promise[T]()
  val firstCompleteHandler = new AtomicReference[Promise[T]](p) with (Try[T] => Unit) {
    override def apply(v1: Try[T]): Unit = getAndSet(null) match {
      case null => ()
      case some => some tryComplete v1
    }
  }
  futures.foreach{ _.filter(condition).onComplete(firstCompleteHandler) }
  p.future
}
于 2018-12-13T20:10:29.483 回答