抛出异常时EDT是否重启？

Question

(下面的示例代码是自包含且可运行的,您可以尝试它,它不会崩溃您的系统:)

Tom Hawtin在这里评论了这个问题:为什么人们在事件队列上运行Java GUI

那:

EDT不太可能崩溃.在EDT调度中抛出的未经检查的异常被捕获,转储并且线程继续.

有人可以解释我这里发生了什么(每次你点击"抛出一个未经检查的异常"按钮,有意地执行除以零):

import javax.swing.*;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.awt.event.WindowAdapter;
import java.awt.event.WindowEvent;

public class CrashEDT extends JFrame {

    public static void main(String[] args) {
        final CrashEDT frame = new CrashEDT();
        frame.addWindowListener(new WindowAdapter() {
            public void windowClosing( WindowEvent e) {
                System.exit(0);
            }
        });
        final JButton jb = new JButton( "throw an unchecked exception" );
        jb.addActionListener( new ActionListener() {
            public void actionPerformed( ActionEvent e ) {
                System.out.println( "Thread ID:" + Thread.currentThread().getId() );
                System.out.println( 0 / Math.abs(0) );
            }
        } );
        frame.add( jb );
        frame.setSize(300, 150);
        frame.setVisible(true);
    }

}

我得到以下消息(这是我期望的):

Exception in thread "AWT-EventQueue-0" java.lang.ArithmeticException: / by zero

对我来说,这是一个未经检查的例外吗？

您可以看到每次触发崩溃时线程ID都会增加.

每次抛出未经检查的异常时,EDT会自动重启,还是未经检查的异常"捕获,转储和线程继续",就像Tom Hawtin评论的那样？

这里发生了什么？

Answer 1

有趣的问题.我本以为抓住了异常并且线程继续进行,但经过一些研究我不太确定.

我用一个扩展你的程序

Set<Thread> seenAwtThreads = new HashSet<Thread>();

我收集了所有"看到"的awt线程,每次单击"throw exception"按钮时,set的大小都会增加,这似乎表明在异常的情况下初始化了一个新线程.

最后我在run执行中发现了这条评论EventDispatchThread:

/*
 * Event dispatch thread dies in case of an uncaught exception. 
 * A new event dispatch thread for this queue will be started
 * only if a new event is posted to it. In case if no more
 * events are posted after this thread died all events that 
 * currently are in the queue will never be dispatched.
 */

完整运行方法的实现如下:

public void run() {
    try {
        pumpEvents(new Conditional() {
            public boolean evaluate() {
                return true;
            }
        });     
    } finally {
        /*
         * This synchronized block is to secure that the event dispatch 
         * thread won't die in the middle of posting a new event to the
         * associated event queue. It is important because we notify
         * that the event dispatch thread is busy after posting a new event
         * to its queue, so the EventQueue.dispatchThread reference must
         * be valid at that point.
         */
        synchronized (theQueue) {
            if (theQueue.getDispatchThread() == this) {
                theQueue.detachDispatchThread();
            }
            /*
             * Event dispatch thread dies in case of an uncaught exception. 
             * A new event dispatch thread for this queue will be started
             * only if a new event is posted to it. In case if no more
             * events are posted after this thread died all events that 
             * currently are in the queue will never be dispatched.
             */
            /*
             * Fix for 4648733. Check both the associated java event
             * queue and the PostEventQueue.
             */
            if (theQueue.peekEvent() != null || 
                !SunToolkit.isPostEventQueueEmpty()) { 
                theQueue.initDispatchThread();
            }
            AWTAutoShutdown.getInstance().notifyThreadFree(this);
        }
    }
}

Answer 2

作为参考，“该机器的特定行为取决于实现。” 例如，线程 ID 在我的平台上保持不变。AWT 线程问题中讨论的最终效果是“当至少有一个可显示组件时，JVM 将不会退出”。