java线程（三）死锁例子

public class TestThread implements Runnable {
	public int flag = 1;
	static Object o1 = new Object();
	static Object o2 = new Object();
	@Override
	public void run() {
		System.out.println("flag" + flag);
		if(flag == 1) {
			synchronized(o1) {
				try {
					Thread.sleep(1000);
				} catch(InterruptedException e) {
					e.printStackTrace();
				}
				synchronized(o2) {
					System.out.println("1");
				}
			}
		}
		
		if(flag == 0) {
			synchronized(o2) {
				try {
					Thread.sleep(1000);
				} catch(InterruptedException e) {
					e.printStackTrace();
				}
				synchronized(o1) {
					System.out.println("0");
				}
			}
		}
	}
	
	
	public static void main(String[] args) {
		TestThread td1 = new TestThread();
		TestThread td2 = new TestThread();
		td1.flag = 1;
		td2.flag = 0;
		Thread t1 = new Thread(td1);
		Thread t2 = new Thread(td2);
		t1.start();
		t2.start();
		
	}
}

线程t1运行的时候他锁定了o1，然后等待sleep1000以后准备锁定o2。可是在他sleep的时候t2已经锁定了o2，同时t2也等待o1.这样就出现了死锁。他们谁也不会得到另一个Object。

面试题的一个例子：

public class TestThread implements Runnable {
	int b = 100;
	
	public synchronized void m1() throws Exception {
		Thread.sleep(5000);
		b = 1000;
		System.out.println("b = " + b);
	}
	
	public void m2() {
		System.out.println(b);
	}
	@Override
	public void run() {
		try {
			m1();
		} catch (Exception e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
	}
	public static void main(String[] args) throws Exception {
		TestThread t1 = new TestThread();
		Thread t = new Thread(t1);
		t.start();
		
		Thread.sleep(1000);
		t1.m2();
	}
	
}

运行结果：
100
b = 1000

从运行结果来看，m2方法被调用了。说明m1被锁定了，但是线程可以访问那些没有被锁定的方法。如果从m2改变b的值也是可以的。