Java面试题2：多线程与synchronized

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本文通过一个简单的并发编程实验探讨了多线程环境下变量自增操作的不可靠性，并提出使用synchronized关键字来解决这一问题的方法。

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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 = " + 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 = " + 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