Scala 创建在构造函数中具有隐式

是否可以创建一个在构造函数中具有隐式的akka参与者？ 具有：

及

我可以使用上下文来创建这样的角色吗

implicit val service:Service = new Service()
val someLong = 3
context.actorOf(FromConfig.props(Props(classOf[Actor], someLong)), "actor")

---

值得一提的是，服务无法传递给构造函数，因为可以接收多个不同的actor类，这些类使用范围中的不同隐式。

在actor的伴生对象中定义您的道具，这有助于注入依赖项

class SomeActor(parameter: Long)(implicit service:Service) extends Actor {
    def receive = {
         case message => // Do your stuff
      }
}

object SomeActor {
   def props(parameter: Long)(implicit service:Service) = Props(new SomeActor(parameter))
}

implicit val service:Service = new Service()
val someLong = 3
val ref = context.actorOf(SomeActor.props(someLong)), "actor")

您可以在此处阅读有关依赖项注入的更多信息：

implicit val service:Service = new Service()
val someLong = 3
context.actorOf(FromConfig.props(Props(classOf[Actor], someLong)), "actor")

class SomeActor(parameter: Long)(implicit service:Service) extends Actor {
    def receive = {
         case message => // Do your stuff
      }
}

object SomeActor {
   def props(parameter: Long)(implicit service:Service) = Props(new SomeActor(parameter))
}

implicit val service:Service = new Service()
val someLong = 3
val ref = context.actorOf(SomeActor.props(someLong)), "actor")