我正在尝试创建一个生产者/消费者类型Scala应用程序。
LoopControl
只是不断向MessageReceiver
发送消息。
MessageReceiver
然后将工作委托给MessageCreatorActor
(其工作是检查地图的对象,如果没有找到则创建一个并启动它)。
此MessageActor
创建的每个MessageCreatorActor
都与ID相关联。
最终,这是我想要做业务逻辑的地方。
但是15分钟后我的内存耗尽了。
任何帮助表示赞赏
import scala.actors.Actor
import java.util.HashMap;
import scala.actors.Actor._
case object LoopControl
case object MessageReceiver
case object MessageActor
case object MessageActorCreator
class MessageReceiver(msg: String) extends Actor {
var messageActorMap = new HashMap[String, MessageActor]
val messageCreatorActor = new MessageActorCreator(null, null)
def act() {
messageCreatorActor.start
loop {
react {
case MessageActor(messageId) =>
if (msg.length() > 0) {
var messageActor = messageActorMap.get(messageId);
if(messageActor == null) {
messageCreatorActor ! MessageActorCreator(messageId, messageActorMap)
}else {
messageActor ! MessageActor
}
}
}
}
}
}
case class MessageActorCreator(msg:String, messageActorMap: HashMap[String, MessageActor]) extends Actor {
def act() {
loop {
react {
case MessageActorCreator(messageId, messageActorMap) =>
if(messageId != null ) {
var messageActor = new MessageActor(messageId);
messageActorMap.put(messageId, messageActor)
println(messageActorMap)
messageActor.start
messageActor ! MessageActor
}
}
}
}
}
class LoopControl(messageReceiver:MessageReceiver) extends Actor {
var count : Int = 0;
def act() {
while (true) {
messageReceiver ! MessageActor ("00-122-0X95-FEC0" + count)
//Thread.sleep(100)
count = count +1;
if(count > 5) {
count = 0;
}
}
}
}
case class MessageActor(msg: String) extends Actor {
def act() {
loop {
react {
case MessageActor =>
println()
println("MessageActor: Got something-> " + msg)
}
}
}
}
object messages extends Application {
val messageReceiver = new MessageReceiver("bootstrap")
val loopControl = new LoopControl(messageReceiver)
messageReceiver.start
loopControl.start
}
答案 0 :(得分:2)
我想知道您的代码是否实际上是在地图中找到现有对象?
如果找不到旧版本,它将继续创建新的MessageActors,直到内存不足为止。
尝试测试。