我如何掌握 Scala Future 中抛出的异常? [英] How do I get hold of exceptions thrown in a Scala Future?

问题描述

我一直在努力回答是否有标准的 Scala 函数来运行带有超时的块?, 如果在 Future 中抛出异常，就会遇到问题.

I've been working up my answer to Is there a standard Scala function for running a block with a timeout?, and have run into a problem if an exception is thrown in a Future.

  def runWithTimeout[T](timeoutMs: Long)(f: => T) : Option[T] = {
    awaitAll(timeoutMs, future(f)).head.asInstanceOf[Option[T]]
  }

这样

runWithTimeout(50) { "result" } should equal (Some("result"))
runWithTimeout(50) { Thread.sleep(100); "result" } should equal (None)

但是如果我在我的块中抛出异常，它不会泄漏，而是被吞下 - 因此以下失败并显示..没有抛出异常"

But if I throw an exception in my block it doesn't leak, but is swallowed - so that the following fails with "..no exception was thrown"

intercept[Exception] {
    runWithTimeout(50) { throw new Exception("deliberate") }
}.getMessage should equal("deliberate")

Syserr 有一个带有消息的堆栈跟踪

Syserr has a stack trace with the message

<function0>: caught java.lang.Exception: deliberate

但我在 Scala 运行时中找不到打印的位置.

but I can't find where in the Scala runtime that is printed.

除了将 f 包装在另一个块中以捕获异常并在抛出时传播它们，还有什么方法可以说服 awaitAll 和/或 Future 抛出?

Apart from wrapping f in another block which catches exceptions and propagates them if thrown, is there any way to persuade awaitAll and/or Future to throw?

推荐答案

简短回答:否.

当您在线程上下文中工作时，异常不会做您想做的事，因为您想知道调用者中的异常，而异常发生在未来的线程中.

Exceptions don't do what you want when you're working in a threaded context, because you want to know about the exception in the caller, and the exception happens in the future's thread.

相反，如果你想知道异常是什么，你应该返回一个Either[Exception,WhatYouWant]——当然，你必须在未来捕捉到那个异常并打包.

Instead, if you want to know what the exception was, you should return an Either[Exception,WhatYouWant]--of course, you have to catch that exception within the future and package it up.

scala> scala.actors.Futures.future{
  try { Right("fail".toInt) } catch { case e: Exception => Left(e) }
}
res0: scala.actors.Future[Product with Serializable with Either[Exception,Int]] = <function0>

scala> res0()   // Apply the future
res1: Product with Serializable with Either[Exception,Int] =
      Left(java.lang.NumberFormatException: For input string: "fail")

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