Question

以下是我正在处理的演员接收部分的示例：

def receive = {
        case "begin" =>
            val listOfFutures: IndexedSeq[Future[Any]] = workers.map(worker => worker ? Work("test"))
            val future: Future[IndexedSeq[Any]] = Future.sequence(listOfFutures)

            future onComplete {
                case Success(result) => println("Eventual result: "+result)
                case Failure(ex) =>  println("Failure: "+ex.getMessage)
            }
        case msg => println("A message received: "+msg)
    }

当其中一个工人的询问失败时（如果发生超时），序列未来将以失败告终。但是我想知道哪些工人失败了。有没有更优雅的方式，而不是简单地逐个映射listOfFutures而不使用 Future.sequence ？

Answer 1

您可以使用未来的recover方法来映射或包装基础异常：

import scala.concurrent.{Future, ExecutionContext}

case class WorkerFailed(name: String, cause: Throwable) 
  extends Exception(s"$name - ${cause.getMessage}", cause)

def mark[A](name: String, f: Future[A]): Future[A] = f.recover {
  case ex => throw WorkerFailed(name, ex)
}

import ExecutionContext.Implicits.global

val f = (0 to 10).map(i => mark(s"i = $i", Future { i / i }))
val g = Future.sequence(f)

g.value  // WorkerFailed: i = 0 - / by zero

Answer 2

感谢@O__我已经提供了另一种可能更适合某些情况的解决方案。

case class WorkerDone(name: String)
case class WorkerFailed(name: String)

import ExecutionContext.Implicits.global

val f = (0 to 10).map {
    i => Future {i/i; WorkerDone(s"worker$i")}.recover{
        case ex => WorkerFailed(s"worker$i")
    }
}
val futureSeq = Future.sequence(f)

futureSeq onComplete {
        case Success(list) => list.collect {case result:WorkerFailed => result}.foreach {failed => println("Failed: "+failed.name)}
        case Failure(ex) => println("Exception: "+ex.getMessage)
    }

// just to make sure program doesn't end before onComplete is called.
Thread.sleep(2000L)

我不确定如果我的例子是一个好习惯，但我的目的是知道哪些工人失败了，无论他们如何失败。

在做Future.sequence时如何知道哪个Future失败了？

2 个答案: