Scala 参与者与非参与者交互（或将消息从参与者同步到 servlet）

发布于 2024-09-12 15:17:36 字数 1277 浏览 5 评论 0原文

我有以下 scala 代码：

  package dummy
  import javax.servlet.http.{HttpServlet,
    HttpServletRequest => HSReq, HttpServletResponse => HSResp}
  import scala.actors.Actor

  class DummyServlet extends HttpServlet {
    RNG.start
    override def doGet(req: HSReq, resp: HSResp) = {
      def message = <HTML><HEAD><TITLE>RandomNumber </TITLE></HEAD><BODY>
           Random number = {getRandom}</BODY></HTML>
      resp.getWriter().print(message)
      def getRandom: String = {var d = new DummyActor;d.start;d.getRandom}
    }
    class DummyActor extends Actor {
      var result = "0"
      def act = { RNG ! GetRandom
        react { case (r:Int) => result = r.toString }
      }
      def getRandom:String = {
        Thread.sleep(300)
        result
      }
    }
  }

  // below code is not modifiable. I am using it as a library
  case object GetRandom
  object RNG extends Actor {
    def act{loop{react{case GetRandom=>sender!scala.util.Random.nextInt}}}
  }

在上面的代码中，我必须使用 thread.sleep 来确保有足够的时间来更新 result，否则 返回 0。在不使用 thread.sleep 的情况下，有什么更优雅的方法可以做到这一点？我想我必须使用 future，但我无法理解这个概念。我需要确保每个HTTP请求得到一个唯一的随机数（当然，随机数只是为了解释问题）。一些提示或参考将不胜感激。

原文

I have the following scala code:

  package dummy
  import javax.servlet.http.{HttpServlet,
    HttpServletRequest => HSReq, HttpServletResponse => HSResp}
  import scala.actors.Actor

  class DummyServlet extends HttpServlet {
    RNG.start
    override def doGet(req: HSReq, resp: HSResp) = {
      def message = <HTML><HEAD><TITLE>RandomNumber </TITLE></HEAD><BODY>
           Random number = {getRandom}</BODY></HTML>
      resp.getWriter().print(message)
      def getRandom: String = {var d = new DummyActor;d.start;d.getRandom}
    }
    class DummyActor extends Actor {
      var result = "0"
      def act = { RNG ! GetRandom
        react { case (r:Int) => result = r.toString }
      }
      def getRandom:String = {
        Thread.sleep(300)
        result
      }
    }
  }

  // below code is not modifiable. I am using it as a library
  case object GetRandom
  object RNG extends Actor {
    def act{loop{react{case GetRandom=>sender!scala.util.Random.nextInt}}}
  }

In the above code, I have to use thread.sleep to ensure that there is enough time for result to get updated, otherwise 0 is returned. What is a more elegant way of doing this without using thread.sleep? I think I have to use futures but I cannot get my head around the concept. I need to ensure that each HTTP reaquest gets a unique random number (of course, the random number is just to explain the problem). Some hints or references would be appreciated.

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天生の放荡 2024-09-19 15:17:36

要么使用：

!! <-- 返回一个你可以等待的 Future

或者

!? <-- 使用带有超时的，完全同步是危险的

鉴于您对 RNG 的定义，这里有一些 REPL 代码来验证：

scala> def foo = { println(RNG.!?(1000,GetRandom)) } 
foo: Unit

scala> foo
Some(-1025916420)

scala> foo
Some(-1689041124)

scala> foo
Some(-1633665186)

文档在这里：http://www.scala-lang.org/api/current/scala/actors/Actor.html

Either use:

!! <-- Returns a Future that you can wait for

!? <-- Use the one with a timeout, the totally synchronous is dangerous

Given your definition of RNG, heres some REPL code to verify:

scala> def foo = { println(RNG.!?(1000,GetRandom)) } 
foo: Unit

scala> foo
Some(-1025916420)

scala> foo
Some(-1689041124)

scala> foo
Some(-1633665186)

Docs are here: http://www.scala-lang.org/api/current/scala/actors/Actor.html

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