①可以有返回值;
②可以抛出异常;
③方法不同:call()方法(Runnable是run()方法);
new Thread().start();
那可怎么办?
package com.zibo.callable;
import java.util.concurrent.Callable;
import java.util.concurrent.ExecutionException;
import java.util.concurrent.FutureTask;
//测试Callable
public class TestCallable {
public static void main(String[] args) throws ExecutionException, InterruptedException {
MyThread myThread = new MyThread();
FutureTask futureTask = new FutureTask(myThread);
new Thread(futureTask,"A").start();
new Thread(futureTask,"B").start();//结果会被缓存,提升效率
//futureTask.get()获取callable的返回值
System.out.println(futureTask.get());//这个get方法会被阻塞,放到最后
}
}
class MyThread implements Callable<String>{
@Override
public String call() throws Exception {
System.out.println("call()执行了!" + Thread.currentThread().getName());
return "123";
}
}
call()执行了!A
123
有缓存;结果可能需要等待,会阻塞;
package com.zibo.add;
import java.util.concurrent.CountDownLatch;
//线程计数器
public class TestCountDownLatch {
public static void main(String[] args) throws InterruptedException {
//创建线程计数器,初始化值为6,也就是说,倒数6条线程,走完结束
CountDownLatch countDownLatch = new CountDownLatch(6);
for (int i = 0; i < 6; i++) {
new Thread(()->{
System.out.println(Thread.currentThread().getName() + "线程执行完了!");
//线程执行完毕,线程计数器减1
countDownLatch.countDown();
},String.valueOf(i)).start();
}
countDownLatch.await();//等待计数器归零,然后向下执行
System.out.println("6个线程执行完毕!");
}
}
0线程执行完了!
4线程执行完了!
3线程执行完了!
2线程执行完了!
1线程执行完了!
5线程执行完了!
6个线程执行完毕!
countDown():使得计数器数量减1;
await():当计数器为0,继续执行下面的方法;
(集齐7颗龙珠,召唤神龙!)
package com.zibo.add;
import java.util.concurrent.BrokenBarrierException;
import java.util.concurrent.CyclicBarrier;
//“加法”计数器
public class TestCyclicBarrier {
public static void main(String[] args) {
//创建计数器,参数1是计数数量,参数2是达到数量时执行的线程
CyclicBarrier cyclicBarrier = new CyclicBarrier(7, () -> {
System.out.println("召唤神龙!");
});
//走7条线程
for (int i = 0; i < 7; i++) {
int finalI = i;
new Thread(()->{
System.out.println(Thread.currentThread().getName() + "搜集了第" + finalI + "颗龙珠!");
try {
cyclicBarrier.await();//等待
} catch (InterruptedException | BrokenBarrierException e) {
e.printStackTrace();
}
},String.valueOf(i)).start();
}
}
}
0搜集了第0颗龙珠!
4搜集了第4颗龙珠!
6搜集了第6颗龙珠!
3搜集了第3颗龙珠!
2搜集了第2颗龙珠!
1搜集了第1颗龙珠!
5搜集了第5颗龙珠!
召唤神龙!
(抢车位:6辆车,3个车位)
package com.zibo.add;
import java.util.concurrent.Semaphore;
//信号量
public class TestSemaphore {
public static void main(String[] args) {
//线程数量,停车位3个
Semaphore semaphore = new Semaphore(3);
//6个线程执行,6辆车
for (int i = 0; i < 6; i++) {
new Thread(()->{
//acquire():从信号量获取许可证
try {
semaphore.acquire();
System.out.println(Thread.currentThread().getName() + "线程拿到许可证,抢到车位,停车!");
System.out.println(Thread.currentThread().getName() + "线程释放许可证,开车离开!");
} catch (InterruptedException e) {
e.printStackTrace();
}finally {
//release():释放许可证,返回到信号量
semaphore.release();
}
},String.valueOf(i)).start();
}
}
}
0线程拿到许可证,抢到车位,停车!
2线程拿到许可证,抢到车位,停车!
2线程释放许可证,开车离开!
1线程拿到许可证,抢到车位,停车!
1线程释放许可证,开车离开!
0线程释放许可证,开车离开!
3线程拿到许可证,抢到车位,停车!
3线程释放许可证,开车离开!
5线程拿到许可证,抢到车位,停车!
5线程释放许可证,开车离开!
4线程拿到许可证,抢到车位,停车!
4线程释放许可证,开车离开!
从信号量获取许可证(业务代码执行前获取);
释放许可证到信号量(业务代码执行后释放);