java例程练习(关于线程同步的补充)

/*
 * 从运行结果看,当m1()方法被锁定后,m2()方法仍然可以执行。
 * 而且b的值被改变。由此可以得出结论:
 * sychronized 只是防止其定义的代码段被同时调用。
 * 
 */


public class Test implements Runnable{
	int b = 100;
	
	public synchronized void m1() throws Exception {
		b = 1000;
		Thread.sleep(5000);
		System.out.println("b = " + b);
	}
	
	public void m2() {
		System.out.println(b);
	}
	
	public void run() {
		try {
			m1();
		} catch(Exception e) {
			e.printStackTrace();
		}
	}
	
	public static void main(String[] args) throws Exception {
		Test t = new Test();
		Thread th = new Thread(t);
		th.start();
		
		Thread.sleep(1000);//确保线程启动
		t.m2();
	}
	
}
/*
运行结果:

	1000
	b = 1000

*/
/*
 * 从运行结果看,当m1()方法被锁定后,m2()方法仍然可以执行。
 * 而且b的值被改变。由此可以得出结论:
 * sychronized 只是防止其定义的代码段被同时调用。
 * 将m2()锁定后,更改部分代码结果???
 * 
 */


public class Test implements Runnable{
	int b = 100;
	
	public synchronized void m1() throws Exception {
		b = 1000;
		Thread.sleep(5000);
		System.out.println("b = " + b);
	}
	
	public synchronized void m2() throws Exception {
		Thread.sleep(2500);
		b = 2000;
	}
	
	public void run() {
		try {
			m1();
		} catch(Exception e) {
			e.printStackTrace();
		}
	}
	
	public static void main(String[] args) throws Exception {
		Test t = new Test();
		Thread th = new Thread(t);
		th.start();
		
		Thread.sleep(1000);//确保线程启动
		t.m2();
		System.out.println(t.b);
	}
	
}
/*
运行结果:

	b = 1000
	2000

*/


posted on 2012-05-05 20:53  Yours风之恋  阅读(114)  评论(0编辑  收藏  举报