Question

我有一个程序可以创建一个actor，然后从默认输入中读取。如果我写下一个特性使得基本的演员在“行动”方法上工作：

 trait SocketActor extends Actor{


protected def sock:Socket


protected val in:BufferedReader=new BufferedReader(new InputStreamReader(this.sock.getInputStream()))
protected val out:PrintWriter= new PrintWriter(this.sock.getOutputStream(), true)

def act(){
     println("This get to be executed")
 }

如果我写下以下内容，则不会执行act方法

 trait SocketActor extends Actor{


protected def sock:Socket


protected val in:ObjectInputStream=new ObjectInputStream(this.sock.getInputStream())
protected val out:ObjectOutputStream = new ObjectOutputStream (this.sock.getOutputStream())
def act(){
     println("This  doesn't get to be executed")
 }

演员的创作可以恢复如下：

import java.net._
import java.io._
import scala.io._
import game.io._
class PlayerActor(protected val sock:Socket) extends {

} with SocketActor
object TabuClient{
    def main(args:Array[String]){
    try{
        println("Always exected on both cases")
        val port=1337
        val s = new Socket(InetAddress.getByName("localhost"), port)

        val a=new PlayerActor(s)
        a.start()

        for (line <- io.Source.stdin.getLines){
            a.sendMessage(line)
        }
        s.close()
    }
    catch{
        case e:Throwable=>{
            e.printStackTrace()
        }

    }
}

两种方式都可以编译，但是第二种方法基本上在actor没有抛出异常后开始失败

Answer 1

对对象输入流执行以下操作解决了它，有趣的是它不需要懒惰，如果它只是一个inputStream

trait SocketActor extends Actor{


protected def sock:Socket


protected lazy val in:ObjectInputStream=new ObjectInputStream(this.sock.getInputStream())
protected lazy val out:ObjectOutputStream = new ObjectOutputStream (this.sock.getOutputStream())
def act(){
     println("This  doesn't get to be executed")
 }

Scala ObjectInputStream使程序失败无声

1 个答案: