java线程 从非常简单的说起谢谢 给点列子啊

[mwc5213948]

给个程序 运行看看啊

[feishare]

一个简单的死锁问题！

package cn.fee.arithmetic;
public class TestDeadLock implements Runnable {
	public int t;
	static Object o1 = new Object();
	static Object o2 = new Object();

	public static void main(String[] args) {
		TestDeadLock d1 = new TestDeadLock();
		TestDeadLock d2 = new TestDeadLock();
		d1.t = 0;
		d2.t = 1;
		Thread t1 = new Thread(d1);
		Thread t2 = new Thread(d2);
		t1.start();
		t2.start();
	}

	public void run() {
		System.out.println(t);
		if (t == 1) {
			synchronized (o1) {
				try {
					Thread.sleep(5000);
				} catch (InterruptedException e) {
					e.printStackTrace();
				}

				synchronized (o2) {
					System.out.println("1");
				}
			}
		}
		if (t == 0) {
			synchronized (o2) {
				try {
					Thread.sleep(500);
				} catch (InterruptedException e) {
					e.printStackTrace();
				}

				synchronized (o1) {
					System.out.println("0");
				}
			}
		}
	}

}

[liyuan0714]

上面不错

[doingendlessly]

ublic class ThreadInterview implements Runnable{
	 int b = 10;
	
	public int getB(){
		return b;
	}
	
	public synchronized void m1() throws Exception{
		b = 3000;
		//try{
			Thread.sleep(2000);
		//}catch(Exception e){
			//e.printStackTrace();
		//}
		System.out.println(Thread.currentThread().getName()+" b=" + b);
	}
	
	//public  synchronized void m2() throws Exception{
	public  void m2() throws Exception{
		//b = 2000;
		//try{
			Thread.sleep(1000);
			b = 2000;
		//}catch(Exception e){
			//e.printStackTrace();
		//}
	}
	
	public void run(){
		try{
		m1();
		}catch(Exception e){
			e.printStackTrace();
		}
	}
	
	public static void main(String[] args) throws Exception {
		// TODO Auto-generated method stub
		ThreadInterview TI = new ThreadInterview();
		Thread t = new Thread(TI);
		t.start();
		
		TI.m2();
		System.out.println(Thread.currentThread().getName() + " b=" +TI.b);
	}
	

}

[doingendlessly]

上面是一更改同一变量b的两个方法m1() m2()其中，m1()有锁synchronized,m2()没有。你可以看看运行结果。
还有，在m2()中， b = 2000;在sleep()前和后运行的结果也不一样。你可以看看。我曾经就这个问题提问过，现在还有点不明白哈。。。