Java-JUC（十三）：现在有两个线程同时操作一个整数I，做自增操作，如何实现I的线程安全性？
2024-09-05 18:58:47
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。
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