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社区首页 >问答首页 >如何使用java.util.concurrent挂起线程?

如何使用java.util.concurrent挂起线程?
EN

Stack Overflow用户
提问于 2014-01-04 12:37:32
回答 2查看 840关注 0票数 2

我到处找过了。如何通过代码挂起/暂停它,直到我使用任何java.util.concurrent方法/对象调用它来唤醒它?我有一个简单的线程和run方法:

当我按下按钮,它就会停止,然后开始,但问题是,当我再次启动它时,我会得到异常。我要它播放/暂停,就像在媒体播放器里一样。

代码语言:javascript
复制
Exception in thread "AWT-EventQueue-0" java.lang.IllegalMonitorStateException

完整的工作代码(例外情况除外):

代码语言:javascript
复制
import java.awt.GridLayout;
import java.awt.event.WindowAdapter;
import java.awt.event.WindowEvent;

import javax.swing.JFrame;
import javax.swing.JOptionPane;
import javax.swing.SwingUtilities;

public class Window extends JFrame {
    ThreadPanel leftPanel, rightPanel;
    Thread leftThread, rightThread;

    public Window() {
        super("StopResume");
    }

    public void createGUI() {
        setLayout(new GridLayout());
        add(leftPanel = new ThreadPanel());
        add(rightPanel = new ThreadPanel());
        leftThread = new Thread(leftPanel);
        rightThread = new Thread(rightPanel);
        leftThread.start();
        rightThread.start();
        setSize(800, 600);
        setVisible(true);
        addWindowListener(new WindowAdapter() {
            public void windowClosing(WindowEvent e) {
                int confirmed = JOptionPane.showConfirmDialog(null, "Zamknąć", "Potwierdzenie", JOptionPane.OK_CANCEL_OPTION);
                if (confirmed == JOptionPane.OK_OPTION) {
                    dispose();//tu podmienic kod
                    System.exit(1);
                }
            }
        });
    }

    public static void main(String[] args) {
        SwingUtilities.invokeLater(new Runnable() {

            @Override
            public void run() {
                new Window().createGUI();
            }
        });

    }

}

代码语言:javascript
复制
 import java.awt.BorderLayout;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.text.DateFormat;
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.concurrent.locks.Condition;
import java.util.concurrent.locks.Lock;
import java.util.concurrent.locks.ReentrantLock;

import javax.swing.JButton;
import javax.swing.JPanel;
import javax.swing.JScrollPane;
import javax.swing.JTextArea;

public class ThreadPanel extends JPanel implements Runnable {
    public static final String SUSPENDED = "GO", RUNNING = "SUSPEND";
    JTextArea txt;
    JButton ppButton;
    DateFormat dateFormat;
    Lock lock;
    Condition cond;
    boolean running;

    public ThreadPanel() {
        super();
        createGUI();
        dateFormat = new SimpleDateFormat("yyyy/MM/dd HH:mm:ss");
        lock = new ReentrantLock();
        cond = lock.newCondition();
        running = true;
    }

    public void createGUI() {
        setLayout(new BorderLayout());
        JScrollPane jsp = new JScrollPane(txt = new JTextArea(), JScrollPane.VERTICAL_SCROLLBAR_ALWAYS, JScrollPane.HORIZONTAL_SCROLLBAR_ALWAYS);
        add(jsp, BorderLayout.CENTER);
        add(ppButton = new JButton(RUNNING), BorderLayout.SOUTH);
        ppButton.addActionListener(new ActionListener() {
            @Override
            public void actionPerformed(ActionEvent arg0) {

                System.out.println(1);
                if (running) {
                    running = false;
                    ppButton.setText(SUSPENDED);
                } else {
                    running = true;
                    ppButton.setText(RUNNING);
                    lock.unlock();
                }
                lock.lock();
                if (!running) {
                    cond.signalAll();
                }

                lock.unlock();

            }
        });
    }

    @Override
    public void run() {
        while (true) {
            lock.lock();
            try {
                if (!running)
                    cond.await();
            } catch (Exception e) {
                e.printStackTrace();
            }
            Calendar cal = Calendar.getInstance();
            txt.append("\n" + dateFormat.format(cal.getTime()));
            try {
                Thread.sleep((long) (Math.random() * 1001 + 500));
            } catch (Exception e) {
                e.printStackTrace();
            }
            System.out.println(2);
            lock.unlock();
        }
    }

}
EN

回答 2

Stack Overflow用户

回答已采纳

发布于 2014-01-04 13:24:47

我看到你想要一个按钮来启动和停止线程。因此,基本上您需要在actionPerformed()中获取锁,查找事物的状态,操作状态,告诉等待实体某些更改,然后释放所有锁。

Runnable线程(由于缺少标签)应该保持大部分不变,但应该检查循环中的Condition,以避免signalAll()唤醒,而Condition仍然不符合或错误的情况。(signal()signalAll()不能保证在释放锁后立即同步,因此可能已经发生了对actionPerformed()的2次调用)。

代码语言:javascript
复制
    public void createGUI() {
        setLayout(new BorderLayout());
        JScrollPane jsp = new JScrollPane(txt = new JTextArea(), JScrollPane.VERTICAL_SCROLLBAR_ALWAYS, JScrollPane.HORIZONTAL_SCROLLBAR_ALWAYS);
        add(jsp, BorderLayout.CENTER);
        add(ppButton = new JButton(RUNNING), BorderLayout.SOUTH);
        ppButton.addActionListener(new ActionListener() {
            @Override
            public void actionPerformed(ActionEvent arg0) {
                // This is where we acquire the lock to safely look at the state
                lock.lock();
                System.out.println(1);
                // Manipulate the state
                if (running) {
                    running = false;
                    ppButton.setText(SUSPENDED);
                } else {
                    running = true;
                    ppButton.setText(RUNNING);
                }

                // Signal that this conditional changed (is either true or false now)
                cond.signalAll();
                // Release the lock so other entities can go forward
                lock.unlock();

            }
        });
    }

    @Override
    public void run() {
        while (true) {
            lock.lock();
            try {
                // This should block until this condition is true with a loop
                while (!running)
                    cond.await();
            } catch (Exception e) {
                e.printStackTrace();
            }
            Calendar cal = Calendar.getInstance();
            txt.append("\n" + dateFormat.format(cal.getTime()));
            // No need to sleep()
            System.out.println(2);
            lock.unlock();
        }
    }

}
票数 1
EN

Stack Overflow用户

发布于 2014-01-04 12:58:46

我认为您所要求的一种方法是使用CyclicBarrier (来自java.util.concurrent),将其参数化为两个“各方”。

在屏障上调用await的第一个线程将被挂起/阻塞,直到第二个线程也调用await,此时两个线程都可以继续执行。

下面是一个简单的代码示例:

代码语言:javascript
复制
import java.util.concurrent.CyclicBarrier;

public class Test {

    public static void main(String[] args) {

        // a barrier requiring two threads to call await before
        // any thread can proceed past the barrier
        final CyclicBarrier barrier = new CyclicBarrier(2);

        new Thread(){
            @Override
            public void run() {
                try {
                    // do some stuff
                    System.out.println("in thread, before the barrier");

                    // calling await blocks until two threads
                    // (this one and one other) have called await
                    barrier.await();

                    // do some more stuff
                    System.out.println("in thread, after the barrier");
                } catch (Exception e) {
                    e.printStackTrace();
                }
            }
        }.start();

        try {
            System.out.println("main thread, before barrier");

            // calling await blocks until two threads
            // (this one and one other) have called await
            barrier.await();

            System.out.println("main thread, after barrier");
        } catch (Exception exc) {
            exc.printStackTrace();
        }
    }
}
票数 1
EN
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/20921083

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