public class tt implements Runnable {
public static int shareVar = 0;
public synchronized void run() {
if (shareVar == 0) {
for (int i = 0; i < 10; i++) {
shareVar++;
if (shareVar == 5) {
try {
this.wait(); }
catch (Exception e) {}
}//if
} //for
} //if
if (shareVar != 0) {
System.out.print(Thread.currentThread().getName());
System.out.println(" shareVar = " + shareVar);
this.notify();
}
} //run

public static void main(String[] args) {
Runnable r = new tt();
Thread t1 = new Thread(r, "t1");
Thread t2 = new Thread(r, "t2");
t1.start();
t2.start();
}
}

t1线程最先执行。由于初始状态下shareVar为0，t1将使shareVar连续加1，当shareVar的值为5时，t1调用wait() 方法， t1将处于休息状态，同时释放锁标志。这时t2得到了锁标志开始执行，shareVar的值已经变为5，所以t2直接输出shareVar的值，
然后再调用notify() 方法唤醒t1。t1接着上次休息前的进度继续执行，但是此时静态变量为５而不是０，应该是不会再累加了，为什么会继续累加到十？