Java面试题2：多线程与synchronized

2 有一个静态变量num，初始值为0。现在开了1000个线程，每个线程内循环1000次，每循环对num自加1，问最后的值是大于、等于还是小于1000000？
答：
（1）小于1000000，因为有1000个线程，不同的线程有可能同时访问num，导致自加的次数变少。

import java.util.concurrent.TimeUnit;

public class Test implements Runnable{
    private static int num = 0;

    @Override
    public void run() {
        for(int i = 1; i <=1000; i++){
            num++;
            System.out.println(Thread.currentThread().getName() +  ", num = " + num  );
        }
    }

    public static void main(String[] args) throws InterruptedException {    
        for (int i = 1; i <=1000; i++) {
            Thread thread = new Thread(new Test());
            thread.setName("Thread："+i);
            thread.start();
        }

        try {
            // 等待全部子线程执行完毕
            TimeUnit.SECONDS.sleep(30);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("Finally, num = "

运行结果：
Thread：19, num = 999981
Thread：19, num = 999982
Thread：19, num = 999983
Thread：975, num = 999367
Thread：975, num = 999984
Thread：975, num = 999985
Thread：975, num = 999986
Thread：975, num = 999987
Thread：975, num = 999988
Thread：975, num = 999989
Thread：975, num = 999990
Finally, num = 999990

（2）若要防止此现象，要用static synchronized关键字对数据进行同步保护。

import java.util.concurrent.TimeUnit;

public class Test implements Runnable{
    private static int num = 0;

    static synchronized private void increaseNumber() {
        num++;
    }

    @Override
    public void run() {
        for(int i = 1; i <=1000; i++){
            increaseNumber();
            System.out.println(Thread.currentThread().getName() +  ", num = " + num  );
        }
    }

    public static void main(String[] args) throws InterruptedException {    
        for (int i = 1; i <=1000; i++) {
            Thread thread = new Thread(new Test());
            thread.setName("Thread："+i);
            thread.start();
        }

        try {
            // 等待全部子线程执行完毕
            TimeUnit.SECONDS.sleep(30);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("Finally, num = "

运行结果：
Thread：3, num = 999993
Thread：3, num = 999994
Thread：3, num = 999995
Thread：3, num = 999996
Thread：3, num = 999997
Thread：3, num = 999998
Thread：3, num = 999999
Thread：3, num = 1000000
Thread：788, num = 999985
Finally, num = 1000000