Question

我正在学习Scala＆amp; Akka并为它开发测试应用程序。在这个应用程序中，几乎所有actor都记录未处理的消息以便于调试：

import akka.actor._

class TestActor extends Actor with ActorLogging {
  def receive: Receive = {
    case Some(value) => // do something...
    case msg => log.debug("Unhandled message: {}.", msg)
  }
}

正如我所说，这些代码几乎存在于我的所有代理中，我开始考虑将其转化为特征：

trait LogUnhandled { this: Actor with ActorLogging =>
  def logUnhandled: Receive = {
    case msg => log.debug("Unhandled message: {}.", msg)
  }
}

并在

之后使用它

class TestActor extends Actor with ActorLogging with LogUnhandled {
  def receive: Receive = {
    case Some(value) => // do something...
  } orElse logUnhandled
}

我很想知道结构子类型是否允许我这样做，或者TestActor#Receive和LogUnhadled#Receive会是不同的类型，但即使在此之前我也有

error: missing parameter type for expanded function
The argument types of an anonymous function must be fully known. (SLS 8.5)
Expected type was: ?

现在除了将第一个{...}块移动到单独的函数中之外，我无法想象如何避免这种情况：

class TestActor extends Actor with ActorLogging with LogUnhandled {
  def doReceive: Receive = {
    case Some(value) => // do something...
  } 

  def receive: Receive = doReceive orElse logUnhandled
}

当然，后者会做，但它有点“错过了重点”并造成其他一些副作用，比如'考虑适当的函数名而不是receive'......

所以我想知道：是否可以通过某种方式声明(Any) => Unit函数签名'来避免'缺少类型参数错误'？

Answer 1

class TestActor extends Actor with ActorLogging with LogUnhandled {
  def receive = ({
    case Some(value) => // do something...
  }: Receive) orElse logUnhandled
}

另请查看this。 LoggingReceive允许您执行以下操作：

class A extends Actor {
   def receive = LoggingReceive {
      case msg => ...
   }
}

所有接受/拒绝的消息都将发布到调试流中。

链接部分函数时如何避免“丢失类型”错误

1 个答案: