根据b站UP主狂神说JUC课程所写的个人学习笔记
视频地址:https://www.bilibili.com/video/BV1B7411L7tE?from=search&seid=14761503393031794075
8.常用的辅助类
8.1CountDownLatch
//计数器
public class CountDownLatchDemo {
public static void main(String[] args) throws InterruptedException {
//倒计时总数是6,必须要执行任务的时候再使用
CountDownLatch countDownLatch = new CountDownLatch(6);
for (int i = 0; i < 6; i++) {
new Thread(()->{
System.out.println(Thread.currentThread().getName()+" go");
countDownLatch.countDown();//数量-1
},String.valueOf(i)).start();
}
countDownLatch.await();//等待计数器归0再向下执行
System.out.println("close door");
}
}
原理:
countDownLatch.countDown();//数量-1
countDownLatch.await();//等待计数器归0再向下执行
每次有线程调用countDown数量-1,假设计数器变为0,await方法就会被唤醒继续执行
8.2 CyclicBarrier
加法计数器
public class CyclicBarrierDemo {
public static void main(String[] args) {
//集齐七颗龙珠召唤神龙
//召唤龙珠的线程
CyclicBarrier cyclicBarrier = new CyclicBarrier(7,()->{
System.out.println("召唤成功");
});
for (int i = 0; i < 7; i++) {
//lambda能操作i吗
final int temp = i;
new Thread(()->{
System.out.println(Thread.currentThread().getName()+"收集"+temp+"龙珠");
try {
cyclicBarrier.await();
} catch (InterruptedException e) {
e.printStackTrace();
} catch (BrokenBarrierException e) {
e.printStackTrace();
}
}).start();
}
}
}
8.3 Semaphore 信号量
抢车位
6个车 3个停车位置
public class SemaphoreDemo {
public static void main(String[] args) {
//线程数量:停车位,限流
Semaphore semaphore = new Semaphore(3);
for (int i = 0; i < 6; i++) {
new Thread(()->{
try {
semaphore.acquire();//得到停车位
System.out.println(Thread.currentThread().getName()+"抢到车位");
TimeUnit.SECONDS.sleep(2);
System.out.println(Thread.currentThread().getName()+"离开车位");
} catch (InterruptedException e) {
e.printStackTrace();
}finally {
semaphore.release();
}
},String.valueOf(i)).start();
}
}
}
原理:
acquire()获得,假设已经满了就等待被释放为止
release()释放,将当前的信号量释放,然后唤醒等待的线程
作用:多个共享资源互斥的使用,并发限流,控制最大的线程数