在这个例子中,我有一个简单的JFrame包含一个绑定了ActionListener的JButton.这个AcitonListener只是更改了一个允许程序完成的布尔标志.
public class Test {
public static void main(String[] args){
final boolean[] flag = new boolean[1];
flag[0] = false;
JFrame myFrame = new JFrame("Test");
JButton myButton = new JButton("Click Me!");
myButton.addActionListener(new ActionListener(){
@Override
public void actionPerformed(ActionEvent arg0) {
System.out.println("Button was clicked!");
flag[0] = true;
}
});
myFrame.add(myButton);
myFrame.setSize(128,128);
myFrame.setVisible(true);
System.out.println("Waiting");
while(!flag[0]){}
System.out.println("Finished");
}
}
这永远不会打印“完成”,并且在点击按钮后打印一次
Waiting Button was clicked!
但是,如果我修改while循环来读取
while(!flag[0]){
System.out.println("I should do nothing. I am just a print statement.");
}
这有效！打印输出看起来像
Waiting I should do nothing. I am just a print statement. I should do nothing. I am just a print statement. .... I should do nothing. I am just a print statement. Button was clicked! Finished
我理解这可能不是等待某个动作的正确方法,但我仍然有兴趣知道为什么Java会以这种方式运行.
解决方法
最有可能的原因是flag [0] = true;在UI线程上执行,而while(！flag [0])在主线程上执行.
如果没有同步,则无法保证UI线程中所做的更改将从主线程中可见.
通过添加System.out.println,您将引入同步点(因为println方法已同步)并且问题得到解决.
您可以将标志设置为易失性实例或类布尔变量(不是数组),或者更简单地,将要执行的任何代码放在侦听器本身中按下的按钮上.
作为参考,带有volatile变量的代码如下所示：
private static volatile boolean flag;
public static void main(String[] args) {
JFrame myFrame = new JFrame("Test");
JButton myButton = new JButton("Click Me!");
myButton.addActionListener(new ActionListener() {
@Override public void actionPerformed(ActionEvent arg0) {
System.out.println("Button was clicked!");
flag = true;
}
});
myFrame.add(myButton);
myFrame.setSize(128,128);
myFrame.setVisible(true);
System.out.println("Waiting");
while (!flag) { }
System.out.println("Finished");
}
