多线程同步问题：Java中CountDownLatch、Semaphore用法示例-蒲公英云

多线程同步问题：Java中CountDownLatch、Semaphore用法示例

原创心已赠人 2024-09-23 07:42 173阅读 0赞

CountDownLatch和Semaphore是Java中两种常用的线程同步工具，主要用于控制多个线程的执行顺序。

CountDownLatch示例：
```java
import java.util.concurrent.CountDownLatch;

public class CountDownLatchExample {
private CountDownLatch countDownLatch = new CountDownLatch(3);

public void threadAction() throws InterruptedException {
    System.out.println("Thread-1 starting...");
    countDownLatch.await();
    System.out.println("Thread-1 completed!");
    // Thread 2 and 3 same process
    System.out.println("Thread-2 starting...");
    countDownLatch.await();
    System.out.println("Thread-2 completed!");
    System.out.println("Thread-3 starting...");
    countDownLatch.await();
    System.out.println("Thread-3 completed!");
}
public static void main(String[] args) {
    CountDownLatchExample example = new CountDownLatchExample();
    Thread thread1 = new Thread(example::threadAction));
    Thread thread2 = new Thread(example::threadAction));
    Thread thread3 = new Thread(example::threadAction));
    thread1.start();
    thread2.start();
    thread3.start();
}

}

在这个例子中，我们创建了一个`CountDownLatch`实例，初始计数为3。然后在3个线程中调用`await()`方法，来等待计数器下降到0。
2. `Semaphore`示例：
```java
import java.util.concurrent.Semaphore;
public class SemaphoreExample {
    private Semaphore semaphore = new Semaphore(5); // 初始化允许的最大并发数量为5
    public void threadAction() {
        try {
            semaphore.acquire(); // 请求一个许可证
            System.out.println("Thread-1 starting...");
            // 线程执行逻辑
            Thread.sleep(2000);
            System.out.println("Thread-1 completed!");
        } catch (InterruptedException e) {
            e.printStackTrace();
        } finally {
            semaphore.release(); // 释放一个许可证
        }
    }
    public static void main(String[] args) {
        SemaphoreExample example = new SemaphoreExample();
        Thread thread1 = new Thread(example::threadAction));
        Thread thread2 = new Thread(example::threadAction));
        thread1.start();
        thread2.start();
    }
}