Che*_*rry 16 debugging scala akka
我创建了akka系统.假设其中有一些演员.我如何用他们的路径打印akka系统中的所有演员?(用于调试目的)
Chr*_*tin 14
Roland Kuhn的回复表明,这不是一个完全无关紧要的问题,但是你可以使用所有参与者都遵守的Identify
-ActorIdentity
请求 - 响应协议,非常接近(对于将在合理时间内回复消息的演员).
一些未经测试的代码拼凑在一起以说明这个想法:
import akka.actor._
def receive = {
case 'listActors =>
context.actorSelection("/user/*") ! Identify()
case path: ActorPath =>
context.actorSelection(path / "*") ! Identify()
case ActorIdentity(_, Some(ref)) =>
log.info("Got actor " + ref.path.toString)
self ! ref.path
}
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ActorSystem
有私有方法printTree
,可用于调试.
1)私人方法呼叫者(来自https://gist.github.com/jorgeortiz85/908035):
class PrivateMethodCaller(x: AnyRef, methodName: String) {
def apply(_args: Any*): Any = {
val args = _args.map(_.asInstanceOf[AnyRef])
def _parents: Stream[Class[_]] = Stream(x.getClass) #::: _parents.map(_.getSuperclass)
val parents = _parents.takeWhile(_ != null).toList
val methods = parents.flatMap(_.getDeclaredMethods)
val method = methods.find(_.getName == methodName).getOrElse(throw new IllegalArgumentException("Method " + methodName + " not found"))
method.setAccessible(true)
method.invoke(x, args: _*)
}
}
class PrivateMethodExposer(x: AnyRef) {
def apply(method: scala.Symbol): PrivateMethodCaller = new PrivateMethodCaller(x, method.name)
}
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2)用法
val res = new PrivateMethodExposer(system)('printTree)()
println(res)
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将打印:
-> / LocalActorRefProvider$$anon$1 class akka.actor.LocalActorRefProvider$Guardian status=0 2 children
?-> system LocalActorRef class akka.actor.LocalActorRefProvider$SystemGuardian status=0 3 children
| ?-> deadLetterListener RepointableActorRef class akka.event.DeadLetterListener status=0 no children
| ?-> eventStreamUnsubscriber-1 RepointableActorRef class akka.event.EventStreamUnsubscriber status=0 no children
| ?-> log1-Logging$DefaultLogger RepointableActorRef class akka.event.Logging$DefaultLogger status=0 no children
?-> user LocalActorRef class akka.actor.LocalActorRefProvider$Guardian status=0 1 children
...
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当心,它可以导致OOM如果你有很多演员.
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