JComponent.paintImmediately()如何在Java Swing中工作?

Lea*_*ner 11 java swing multithreading jcomponent

我的理解: 与Swing中的大多数组件/操作不同,对JComponent.repaint()的调用是线程安全的,即虽然重新绘制请求来自另一个线程(即不是来自EDT),但实际绘制仅在EDT中进行.下面的代码片段演示了这一点.

public class PaintingDemo {

    public static void main(String[] args) {
        final JFrame frame = new JFrame();
        final JPanel p = new MyPanel();
        SwingUtilities.invokeLater(new Runnable() {
            @Override
            public void run() {
                frame.add(p, BorderLayout.CENTER);
                frame.setSize(200, 200);
                frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
                frame.setVisible(true);
            }
        });
        new Thread("MyThread") {
            public void run() {
                while (true) {
                // Below statements are important to show the difference
                    p.repaint();
                    p.paintImmediately(p.getBounds());
                    try {
                        Thread.sleep(1000);
                    } catch(Exception e) {}
                }
            }
        }.start();
    }

}

class MyPanel extends JPanel {
    @Override
    public void paint(Graphics g) {
        System.out.println("paint() called in "+ Thread.currentThread().getName());
        super.paint(g);
    }
}
Run Code Online (Sandbox Code Playgroud)

从输出中可以看出,当调用repaint()时,绘制是在EDT中完成的,无论调用哪个线程 - 所以没有问题.但是,在paintImmediately()的情况下 - 绘制发生在调用它的同一个线程中.

考虑一种情况,其中EDT正在改变组件的状态,另一个线程(从中调用paintImmediately())正在绘制相同的组件.

我的问题: 在paintImmediately()的情况下,如何处理Event Dispatcher Thread(EDT)和其他线程之间的同步?

Gui*_*let 7

据我所知,当你调用paintImmediately时,你调用以下代码:

        Component c = this;
        Component parent;

        if(!isShowing()) {
            return;
        }

        JComponent paintingOigin = SwingUtilities.getPaintingOrigin(this);
        if (paintingOigin != null) {
            Rectangle rectangle = SwingUtilities.convertRectangle(
                    c, new Rectangle(x, y, w, h), paintingOigin);
            paintingOigin.paintImmediately(rectangle.x, rectangle.y, rectangle.width, rectangle.height);
            return;
        }

        while(!c.isOpaque()) {
            parent = c.getParent();
            if(parent != null) {
                x += c.getX();
                y += c.getY();
                c = parent;
            } else {
                break;
            }

            if(!(c instanceof JComponent)) {
                break;
            }
    }
    if(c instanceof JComponent) {
        ((JComponent)c)._paintImmediately(x,y,w,h);
    } else {
        c.repaint(x,y,w,h);
    }
Run Code Online (Sandbox Code Playgroud)

因此,除非这不是a,否则JComponent你最终会调用_paintImmediately()最终调用,paint(Graphics)如下所示的堆栈跟踪(从我将在本文末尾发布的一段代码中捕获):

Thread [pool-1-thread-1] (Suspended)    
    TestPaint$1.paint(Graphics) line: 23    
    TestPaint$1(JComponent).paintToOffscreen(Graphics, int, int, int, int, int, int) line: 5221 
    RepaintManager$PaintManager.paintDoubleBuffered(JComponent, Image, Graphics, int, int, int, int) line: 1482 
    RepaintManager$PaintManager.paint(JComponent, JComponent, Graphics, int, int, int, int) line: 1413  
    RepaintManager.paint(JComponent, JComponent, Graphics, int, int, int, int) line: 1206   
    TestPaint$1(JComponent)._paintImmediately(int, int, int, int) line: 5169    
    TestPaint$1(JComponent).paintImmediately(int, int, int, int) line: 4980 
    TestPaint$1(JComponent).paintImmediately(Rectangle) line: 4992  
    TestPaint$3.run() line: 50  
    ThreadPoolExecutor.runWorker(ThreadPoolExecutor$Worker) line: 1110  
    ThreadPoolExecutor$Worker.run() line: 603   
    Thread.run() line: 722  
Run Code Online (Sandbox Code Playgroud)

但是如果你试图同时调用repaint()(从另一个Thread),你会看到两个同时运行(我尝试使用debuger踩代码并且绘画从未停止在另一个Thread中出现)似乎在Java代码级别,没有太多同步(至少我找不到任何东西).因此,如果您最终修改了EDT中的组件状态,我相信结果是非常不可预测的,您应该通过各种方式避免这种情况.

为了说明我的观点,我尝试修改paint方法中变量的状态,添加一个sleep来增加2个线程(EDT和其他)之间冲突的风险,并且看起来两个线程之间没有同步(的System.err.println()输出null从时间到时间).

现在我想知道为什么你需要立即执行paint.除非你阻止EDT,否则没有太多合理的理由来执行这样的事情.

下面是我用来测试这些东西的代码(非常接近问题中发布的代码).代码只是试图了解正在发生的事情,而不是展示如何进行正确的绘画或任何良好的Swing练习.

import java.awt.Color;
import java.awt.Graphics;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.Random;
import java.util.concurrent.Executors;

import javax.swing.JFrame;
import javax.swing.JPanel;
import javax.swing.SwingUtilities;
import javax.swing.Timer;

public class TestPaint {

    protected void initUI() {
        JFrame frame = new JFrame();
        frame.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE);
        frame.setTitle(TestPaint.class.getSimpleName());
        final Random rand = new Random();
        final JPanel comp = new JPanel() {
            private String value;

            @Override
            public void paint(Graphics g) {
                value = "hello";
                super.paint(g);
                try {
                    Thread.sleep(20);
                } catch (InterruptedException e) {
                    // TODO Auto-generated catch block
                    e.printStackTrace();
                }
                g.setColor(new Color(rand.nextInt(256), rand.nextInt(256), rand.nextInt(256)));
                g.fillRect(0, 0, getWidth(), getHeight());
                if (SwingUtilities.isEventDispatchThread()) {
                    System.err.println("Painting in the EDT " + getValue());
                } else {
                    System.err.println("Not painting in EDT " + getValue());
                }
                value = null;
            }

            public String getValue() {
                return value;
            }
        };
        frame.add(comp);
        frame.setSize(400, 400);
        frame.setLocationRelativeTo(null);
        frame.setVisible(true);
        Timer t = new Timer(1, new ActionListener() {

            @Override
            public void actionPerformed(ActionEvent e) {
                comp.repaint();
            }
        });
        t.start();
        Executors.newSingleThreadExecutor().execute(new Runnable() {

            @Override
            public void run() {
                while (true) {
                    comp.paintImmediately(comp.getBounds());
                }
            }
        });
    }

    public static void main(String[] args) {
        SwingUtilities.invokeLater(new Runnable() {
            @Override
            public void run() {
                new TestPaint().initUI();
            }
        });
    }

}
Run Code Online (Sandbox Code Playgroud)

  • 感谢@Guillaume亲自尝试并分享您的研究成果.我同意使用paintImmediately()的原因并不多,但我只是想了解"它是如何工作的?" 因为javadoc说......"如果在调度当前事件时需要更新显示,这个方法很有用." 此外,这种方法是公开的,并且绝对没有任何关于其警告的警告; 至少javadoc是误导性的.现在我明白了paintImmediately()需要在EDT中调用,而且与repaint()不同,它不是线程安全的. (2认同)