问题分析:正如i在多线程中如果想实现i的多线程操作,必须i要使用volitle来保证其内存可见性,但是i++自增操作不具备原子性操作,因此需要对i++这段代码确保其原子性操作即可。
方案1:
使用ReetranLock实现i++的原子性操作。
private static volatile int i=0; public static void main(String[] args) throws InterruptedException { CountDownLatch countDownLatch=new CountDownLatch(2); Lock lock=new ReentrantLock(); Thread thread1 = new Thread(new Runnable() { @Override public void run() { try{ lock.lock(); i++; }finally{ lock.unlock(); countDownLatch.countDown(); } } },"Thread-1"); Thread thread2 = new Thread(new Runnable() { @Override public void run() { try{ lock.lock(); i++; }finally{ lock.unlock(); countDownLatch.countDown(); } } },"Thread-2"); thread1.start(); thread2.start(); countDownLatch.await(); System.out.println(i); }
方案2:
使用Semaphore实现i++的原子性操作。
private static volatile int i = 0; public static void main(String[] args) throws InterruptedException { CountDownLatch countDownLatch = new CountDownLatch(2); Semaphore semaphore = new Semaphore(1); Thread thread1 = new Thread(new Runnable() { @Override public void run() { try { semaphore.acquire(); } catch (InterruptedException e) { e.printStackTrace(); } i++; semaphore.release(); countDownLatch.countDown(); } }, "Thread-1"); Thread thread2 = new Thread(new Runnable() { @Override public void run() { try { semaphore.acquire(); } catch (InterruptedException e) { e.printStackTrace(); } i++; semaphore.release(); countDownLatch.countDown(); } }, "Thread-2"); thread1.start(); thread2.start(); countDownLatch.await(); System.out.println(i); }
当然也可以选择sychronized方式实现。