为什么我的Scala未来无法传播错误？

Question

This code works as expected:

it("should propagate exceptions") {
  intercept[RuntimeException] {
    val future = Future { Thread.sleep(10); sys.error("whoops"); 22 }
    Await.result(future, Duration.Inf)
  }.getMessage should equal ("whoops")
}

But this doesn't:

it("should propagate errors") {
  intercept[StackOverflowError] {
    val future = Future { Thread.sleep(10); throw new StackOverflowError("dang"); 22 }
    Await.result(future, Duration.Inf)
  }.getMessage should equal ("dang")
}

The future in this second test never returns. Why doesn't an Error subclass (as opposed to an Exception subclass) terminate my future? How should I handle Error s?

EDIT: This is possibly related, but not identical, to Why does Scala Try not catching java.lang.StackOverflowError? . I'm not using Try here. The core issue is that the Future never returns at all; I can't catch any error from it because it just hangs.

Answer 1

As pointed out in the comments, this is a duplicate of Why does Scala Try not catching java.lang.StackOverflowError?

According to Scala documentation.

Note: only non-fatal exceptions are caught by the combinators on Try (see >scala.util.control.NonFatal). Serious system errors, on the other hand, will be >thrown.

No Throwable -> Errors are catched by Try

Also to answer your question about how error handling is usually done. In Scala you can use try / catch for code that can cause exceptions (very similar to Java):

try { 
// ... Your dangerous code in here
} catch {
 case ioe: IOException => ... //
 case e: Exception => ...
} 

And you should always have the more specific exceptions first.

The code you provided would look something like this: https://scastie.scala-lang.org/2DJXJ6ESS9ySJZSwSodmZg

Also I tried out your code and it definitely produces the StackOverFlowerror. But it can't catch it properly like the above mentioned link explains.

Answer 2

The reporter facility is for catastrophes, which just hooks into the thread's UncaughtExceptionHandler , but it looks like it works out of the box with just the default thread factory:

scala 2.13.0-M5> import concurrent._,java.util.concurrent.Executors
import concurrent._
import java.util.concurrent.Executors

scala 2.13.0-M5> val ec = ExecutionContext.fromExecutor(null, e => println(s"Handle: $e"))
ec: scala.concurrent.ExecutionContextExecutor = scala.concurrent.impl.ExecutionContextImpl$$anon$3@5e7c141d[Running, parallelism = 4, size = 0, active = 0, running = 0, steals = 0, tasks = 0, submissions = 0]

scala 2.13.0-M5> val f = Future[Int](throw new NullPointerException)(ec)
f: scala.concurrent.Future[Int] = Future(<not completed>)

scala 2.13.0-M5> f
res0: scala.concurrent.Future[Int] = Future(Failure(java.lang.NullPointerException))

scala 2.13.0-M5> val f = Future[Int](throw new StackOverflowError)(ec)
Handle: java.lang.StackOverflowError
f: scala.concurrent.Future[Int] = Future(<not completed>)

whereas

scala 2.13.0-M5> val ec = ExecutionContext.fromExecutor(Executors.newSingleThreadExecutor, e => println(s"Handle: $e"))
ec: scala.concurrent.ExecutionContextExecutor = scala.concurrent.impl.ExecutionContextImpl@317a118b

scala 2.13.0-M5> val f = Future[Int](throw new StackOverflowError)(ec)
f: scala.concurrent.Future[Int] = Future(<not completed>)
Exception in thread "pool-1-thread-1" java.lang.StackOverflowError
    at $line14.$read$$iw$$iw$$iw$$iw$.$anonfun$f$1(<console>:1)
    at scala.concurrent.Future$.$anonfun$apply$1(Future.scala:659)
    at scala.util.Success.$anonfun$map$1(Try.scala:261)
    at scala.util.Success.map(Try.scala:209)
    at scala.concurrent.impl.Promise$Transformation.doMap(Promise.scala:420)
    at scala.concurrent.impl.Promise$Transformation.run(Promise.scala:402)
    at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1149)
    at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:624)
    at java.lang.Thread.run(Thread.java:748)

You could construct a rig that registers a future when it runs, and a safe await that knows when threads have blown up. Maybe you want to retry an algorithm with a lower max recursion depth, for example.