两个线程交替打印1-100之间的奇偶数

实现一:synchronized关键字实现(缺点:效率低)

实现思路:让两个线程来争夺同一把锁,如果抢到锁只有满足自己的奇偶判定条件才打印(注:&方式判断奇偶更高效)

public class Demo04_两个线程交替打印奇数_纯synchronized {
    private static Object lock = new Object();
    private static int i = 1;
    public static void main(String[] args) throws InterruptedException {
        Thread t1 = new Thread(() -> {
            while (i < 100) {
                synchronized (lock) {
                    if ((i & 1) == 1) {
                        System.out.println(i++ + "是奇数");
                    }
                }
            }
        });
        Thread t2 = new Thread(() -> {
            while (i < 100) {
                synchronized (lock) {
                    if ((i & 1) == 0) {
                        System.out.println(i++ + "是偶数");
                    }
                }
            }
        });
        t1.start();;
        t2.start();
    }
}

实现二:wait/notify()(效率高)

思路:进行进程间的通信,奇线程打印完阻塞自己,偶线程打印完唤醒被阻塞的线程,同时将自己阻塞

public class Demo03_两个线程交替打印奇偶数 {
    private static Object lock = new Object();
    private static int i = 1;
    public static void main(String[] args) {
        Thread t1 = new Thread(() -> {
            for (; i <= 100;) {
                synchronized (lock) {
                        System.out.println(i ++ + ":我是偶数");
                        try {
                            lock.notify();
                            if (i <= 100) {
                                lock.wait();
                            }
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                }
            }
        });
        Thread t2 = new Thread(() -> {
            for (; i <= 100;) {
                synchronized (lock) {
                        System.out.println(i ++ + ":我是奇数");
                        try {
                            if (i <= 100) {
                                lock.wait();
                            }
                            lock.notify();
                        } catch (InterruptedException e) {
                            e.printStackTrace();
                        }
                }
            }
        });
        t2.start();
        t1.start();
    }

 

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