等待和通知
等待和通知的标准范式
等待方:
1:获取对象的锁
2:在循环中判断条件是否满足,不满足调用wait方法继续阻塞,为啥要要循环中判断呢?因为该线程被唤醒之后可能条件依旧不满足
3:条件满足,执行业务逻辑
通知方:
1:获取对象的锁
2:改变相关条件
3:通知所有等待在对象的线程
都是属于Object的方法
等待:wait
通知:notify/notifyAll
需求:一个快递在变更;里程数和地点的时候通知等待的线程处理变更后的请求
测试使用notifyAll唤醒
实体类
package org.dance.day1.wn;
/**
* 快递实体类
*
* @author ZYGisComputer
*/
public class Express {
public final static String CITY = "ShangHai";
/**
* 快递运输的里程数
*/
private int km;
/**
* 快递到达的地点
*/
private String site;
public Express() {
}
public Express(int km, String site) {
this.km = km;
this.site = site;
}
/**
* 变化公里数:然后通知处于wait状态并需要处理公里数的线程进行业务处理
*/
public synchronized void checkKm() {
// 变化公里数
this.km = 101;
// 全部通知
notifyAll();
}
/**
* 变化地点:然后通知处于wait状态并需要处理地点的线程进行业务处理
*/
public synchronized void checkSite() {
// 变化城市
this.site = "BeiJin";
// 全部通知
notifyAll();
}
public synchronized void waitKm() {
// 循环等待
while (this.km <= 100) {
try {
wait();
System.out.println("check km " + Thread.currentThread().getId());
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println("the km is " + this.km + ", I will change DB.");
}
public synchronized void waitSite() {
// 循环等待
while (CITY.equals(this.site)) {
try {
wait();
System.out.println("check site " + Thread.currentThread().getId());
} catch (InterruptedException e) {
e.printStackTrace();
}
}
System.out.println("the site is " + this.site + ", I will change DB.");
}
}
测试类
package org.dance.day1.wn;
import org.dance.tools.SleepTools;
/**
* 测试wait notify/notifyAll
* @author ZYGisComputer
*/
public class TestWN {
private static Express express = new Express(0,Express.CITY);
/**
* 检查里程数变化的线程,不满足一直等待
*/
private static class CheckKm extends Thread{
@Override
public void run() {
express.waitKm();
}
}
/**
* 检查城市变化的线程,不满足一直等待
*/
private static class CheckSite extends Thread{
@Override
public void run() {
express.waitSite();
}
}
public static void main(String[] args) {
for (int i = 0; i < 3; i++) {
new CheckKm().start();
}
for (int i = 0; i < 3; i++) {
new CheckSite().start();
}
SleepTools.second(1);
// 修改里程数
express.checkKm();
}
}
测试结果:
check site 16
check site 15
check site 14
check km 13
the km is 101, I will change DB.
check km 12
the km is 101, I will change DB.
check km 11
the km is 101, I will change DB.
测试发现全部的线程全部被唤醒了,然后其中三个等待城市变化的线程再次进入阻塞,另外三个等待里程数变化的执行成功退出阻塞
返回结果:
check km 11
the km is 101, I will change DB.
check site 11
因为notify通知任意一个在这个对象上阻塞的线程,如果正好通知到了,等待里程数的,那么也只有一个被唤醒,其他两个继续阻塞,如果通知到了一个等待城市变化的那么这个线程将继续进入阻塞,所以看来,我们应该尽量使用notifyAll少用notify,因为可能发生信号丢失的情况
/**
* Wakes up a single thread that is waiting on this object's
* monitor. If any threads are waiting on this object, one of them
* is chosen to be awakened. The choice is arbitrary and occurs at
* the discretion of the implementation. A thread waits on an object's
* monitor by calling one of the {@code wait} methods.
* <p>
* The awakened thread will not be able to proceed until the current
* thread relinquishes the lock on this object. The awakened thread will
* compete in the usual manner with any other threads that might be
* actively competing to synchronize on this object; for example, the
* awakened thread enjoys no reliable privilege or disadvantage in being
* the next thread to lock this object.
* <p>
* This method should only be called by a thread that is the owner
* of this object's monitor. A thread becomes the owner of the
* object's monitor in one of three ways:
* <ul>
* <li>By executing a synchronized instance method of that object.
* <li>By executing the body of a {@code synchronized} statement
* that synchronizes on the object.
* <li>For objects of type {@code Class,} by executing a
* synchronized static method of that class.
* </ul>
* <p>
* Only one thread at a time can own an object's monitor.
*
* @throws IllegalMonitorStateException if the current thread is not
* the owner of this object's monitor.
* @see java.lang.Object#notifyAll()
* @see java.lang.Object#wait()
*/
public final native void notify();
在源码中可以看到,这个方法是一个 native的
在他的描述中有一段
The choice is arbitrary and occurs at,the discretion of the implementation.
翻译为中文
选择是任意的,发生在,执行的自由裁量权。
所以说notify的唤醒是随意的,并且信号只发出一次
但是据有人说,在线程进入等待的时候会进入一个等待队列,notify会唤醒第一个等待的线程
我得到的结果就是在HotSpot虚拟机当中 notify唤醒的是阻塞线程队列当中的第一个ObjectWaiter节点,其他虚拟机不一定.
我觉得这个问题也已当做一个在面试的时候,你问面试官的一个技术性问题
作者:彼岸舞
时间:2020\09\16
内容关于:并发编程
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原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。