③. JUC强大的三个工具类
3>.
JUC强大的三个工具类 掌握
为什么这里要介绍下JUC强大的工具类?
CountDownLatch | CyclicBarrier | Semaphore 底层都是AQS来实现的
①. CountDownLatch(闭锁)
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①. CountDownLatch主要有两个方法,当一个或多个线程调用await方法时,这些线程会阻塞
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②. 其它线程调用countDown方法会将计数器减1(调用countDown方法的线程不会阻塞)
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③. 计数器的值变为0时,因await方法阻塞的线程会被唤醒,继续执行
//需求:要求6个线程都执行完了,mian线程最后执行
public class CountDownLatchDemo {
public static void main(String[] args) throws Exception{
CountDownLatch countDownLatch=new CountDownLatch(6);
for (int i = 1; i <=6; i++) {
new Thread(()->{
System.out.println(Thread.currentThread().getName()+"\t");
countDownLatch.countDown();
},i+"").start();
}
countDownLatch.await();
System.out.println(Thread.currentThread().getName()+"\t班长关门走人,main线程是班长");
}
}
- ④. 利用枚举减少if else的判断
public enum CountryEnum {
one(1,"齐"),two(2,"楚"),three(3,"燕"),
four(4,"赵"),five(5,"魏"),six(6,"韩");
private Integer retCode;
private String retMessage;
private CountryEnum(Integer retCode,String retMessage){
this.retCode=retCode;
this.retMessage=retMessage;
}
public static CountryEnum getCountryEnum(Integer index){
CountryEnum[] countryEnums = CountryEnum.values();
for (CountryEnum countryEnum : countryEnums) {
if(countryEnum.getRetCode()==index){
return countryEnum;
}
}
return null;
}
public Integer getRetCode() {
return retCode;
}
public String getRetMessage() {
return retMessage;
}
}
/*
楚 **国,被灭
魏 **国,被灭
赵 **国,被灭
燕 **国,被灭
齐 **国,被灭
韩 **国,被灭
main **秦国一统江湖
* */
public class CountDownLatchDemo {
public static void main(String[] args) throws Exception{
CountDownLatch countDownLatch=new CountDownLatch(6);
for (int i = 1; i <=6; i++) {
new Thread(()->{
System.out.println(Thread.currentThread().getName()+"\t"+"**国,被灭");
countDownLatch.countDown();
},CountryEnum.getCountryEnum(i).getRetMessage()).start();
}
countDownLatch.await();
System.out.println(Thread.currentThread().getName()+"\t"+"**秦国一统江湖");
}
}
②. CyclicBarrier
2>.
CyclicBarrier
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①. CyclicBarrier的字面意思是可循环(Cyclic) 使用的屏障(barrier).它要做的事情是,让一组线程到达一个屏障(也可以叫做同步点)时被阻塞,知道最后一个线程到达屏障时,屏障才会开门,所有被屏障拦截的线程才会继续干活,线程进入屏障通过CyclicBarrier的await()方法
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②. 代码验证:
//集齐7颗龙珠就能召唤神龙
public class CyclicBarrierDemo {
public static void main(String[] args) {
// public CyclicBarrier(int parties, Runnable barrierAction) {}
CyclicBarrier cyclicBarrier=new CyclicBarrier(7,()->{
System.out.println("召唤龙珠");
});
for (int i = 1; i <=7; i++) {
final int temp=i;
new Thread(()->{
System.out.println(Thread.currentThread().getName()+"\t收集到了第"+temp+"颗龙珠");
try {
cyclicBarrier.await();
} catch (InterruptedException e) {
e.printStackTrace();
} catch (BrokenBarrierException e) {
e.printStackTrace();
}
}).start();
}
}
}
③.Semaphore(信号量)
3>.
Semaphore(信号量)
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①. acquire(获取) 当一个线程调用acquire操作时,它要么通过成功获取信号量(信号量减1),要么一直等下去,直到有线程释放信号量,或超时。
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②. release(释放)实际上会将信号量的值加1,然后唤醒等待的线程。
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③. 信号量主要用于两个目的,一个是用于多个共享资源的互斥使用,另一个用于并发线程数的控制。
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④. 代码验证
public class SemaphoreDemo {
public static void main(String[] args) {
Semaphore semaphore=new Semaphore(3);
for (int i = 1; i <=6; i++) {
new Thread(()->{
try {
System.out.println(Thread.currentThread().getName()+"\t抢占了车位");
semaphore.acquire();
System.out.println(Thread.currentThread().getName()+"\t离开了车位");
} catch (InterruptedException e) {
e.printStackTrace();
}finally {
semaphore.release();
}
},String.valueOf(i)).start();
}
}
}
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