如何使在 Java 中阻塞的函数超时？

Question

基本上，我正在调用 BufferedReader.ReadLine();但是我在一个多线程服务器中，我正在同步树中的一个节点。因此，当调用此 ReadLine 函数时，如果其他人到达该节点，他们就会被锁定。我不知道如何对 ReadLine 在退出线程之前等待响应的时间量进行时间限制。我得到的最接近的是创建一个会休眠 1ms 的新线程，然后检查我设置 ReadLine 的变量是否已更改。所以是这样的：

synchronized (pointer) {
    String answer = "";
    Thread d = new Thread(new Runnable() {
        public void run() {
            try {
                int i = 0;
                while (answer.equals("")) {
                    if (i == 10000) {
                        System.out.println("Timeout Occured");
                        System.exit(0);
                    }
                    try {
                        Thread.sleep(1);
                        i++;
                    }
                    catch(Exception e) {
                        System.out.println("sleep problem occured");
                    }
                }
            }
            catch (IOException ex) {
            }
        }    
    });

    d.start();
    answer = socketIn.readLine();
}

这达到了我想要的效果，但我不知道如何停止当前线程以解锁节点，以便其他用户可以继续而不是杀死整个服务器。最后，我想也许我可以这样做：

    Thread d = new Thread(new Runnable() {
        public void run() {
            try {
                answer = socketIn.readLine(); 
            } catch (IOException ex) {
            }
        }    
    });

    d.join(10000);
    catch (InterruptedException e){
    socketOut.println("Timeout Occured. Returning you to the beginning...");
    socketOut.flush();
    return;
}

但这似乎仍然阻塞并且无法继续。有人可以帮我解决这个问题吗？我不明白我做错了什么？

我也试图让 ExecutorService 工作，但不能。这是我的答案吗？我将如何实现它？

[EDIT] socketIn 是一个 BufferedReader，应该明确表示抱歉。 此外，客户端通过 telnet 进行连接，但我认为这并不重要。

我在这里做的是一个“名人猜谜游戏”，用户可以在其中将名人添加到树中。所以我需要锁定这个人正在编辑的节点以保证线程安全

Answer 1

这是作业吗？这与昨天其他人提出的问题非常接近。如果是这样，它应该有作业标签。

只有当一个线程将修改其他线程可能读取/修改的数据时，您才需要锁定某些东西。

如果您要锁定等待输入的内容，则锁定的范围太广了。

您的流程应该是：

• 从客户端读取输入（阻塞 readLine()）
• 锁定共享资源
• 修改
• 解锁

（这是假设您每个连接/客户端都有一个线程，并且正在阻止来自客户端的读取）

话虽如此...如果您正在从套接字读取并希望它超时，则需要在第一次接受连接时使用clientSocket.setSoTimeout(1000);。如果您的 BufferedReader 正在等待该时间（以毫秒为单位）并且没有得到输入，它将抛出 java.net.SocketTimeoutException

String inputLine = null;
try 
{
    inputLine = in.readLine();
    if (inputLine == null)
    {
        System.out.println("Client Disconnected!");
    }
    else 
    {
        // I have input, do something with it
    }
}
catch(java.net.SocketTimeoutException e)
{
    System.out.println("Timed out trying to read from socket");
}

Answer 2

一切都已经完成了。尝试使用java.util.concurrent。

    //
    // 1. construct reading task
    //
    final FutureTask<String> readLine = new FutureTask<String> (
        new Callable<String>() {
            @Override public String call() throws Exception {
                return socketIn.readLine();
            }
        }
    );
    //
    // 2. wrap it with "timed logic"
    //    *** remember: you expose to your users only this task
    //
    final FutureTask<String> timedReadLine = new FutureTask<String> (
        new Callable<String>() {
            @Override public String call() throws Exception {
                try {
                    //
                    // you give reading task a time budget:
                    //      try to get a result for not more than 1 minute
                    //
                    return readLine.get( 1, TimeUnit.MINUTES );
                } finally {
                    //
                    // regardless of the result you MUST interrupt readLine task
                    // otherwise it might run forever
                    //      *** if it is already done nothing bad will happen 
                    //
                    readLine.cancel( true );
                }
            }
        }
    )
    {
        //
        // you may even protect this task from being accidentally interrupted by your users:
        //      in fact it is not their responsibility
        //
        @Override
        public boolean cancel(boolean mayInterruptIfRunning) {
            return false;
        }
    };

    Executor executor = Executors.newCachedThreadPool();

    // 3. execute both
    executor.execute( readLine );
    executor.execute( timedReadLine );

    // 4. ...and here comes one of your users who can wait for only a second
    try {
        String answer = timedReadLine.get(1, TimeUnit.SECONDS);
        //
        // finally user got his (her) answer
        //
    } catch (InterruptedException e) {
        //
        // someone interrupted this thread while it was blocked in timedReadLine.get(1, TimeUnit.SECONDS)
        //
    } catch (ExecutionException e) {
         //
        // e.getCause() probably is an instance of IOException due to I/O failure
        //
    } catch (TimeoutException e) {
        //
        // it wasn't possible to accomplish socketIn.readLine() in 1 second
        //
    }

Answer 3

我想出了一个解决办法：

answer = "";
try{
    Thread d = new Thread(new Runnable() {
        public void run() {
            try {
                answer = socketIn.readLine();
            }   
            catch (IOException ex) {
                System.out.println("IO exception occurred");
            }
        }
    });
    d.join(10000); //Not sure if this is superfluous or not, but it didn't seem to work without it.
    d.start();
    i = 0;
    while (true){
        if (i == 10000){
            if (d.isAlive()) throw InterruptedException;
        }
        if (answer.equals("")){
            Thread.sleep(1);
        }
        else{
            break;
        }
        i++;
    }
    //This essentially acts as Thread.sleep(10000), but the way I 
    //implemented it, it checks to see if answer is modified or not every  
    //.001 seconds. It will run for just over 10 seconds because of these checks. 
    //The number in Thread.sleep could probably be higher, as it is 
    //unnecessary to check so frequently and it will make the code more efficient
    //Once it hits 10000, it throws an exception to move to the catch block below
    //where the catch block returns everything to its original state and 
    //returns the client to the beginning
    }
catch (Exception e){
    socketOut.println("Timeout Occurred. Returning you to the beginning...");
    socketOut.flush();
    pointer = tree.root;
    //reset class members to their original state
    return;
}

感谢您观看，Brian Roach。你的帖子内容丰富，但我不小心遗漏了一些关键信息。以后我会更加小心的。