下面的java代码示例使用java DelayQueue来处理任务。然而,从另一个线程插入任务似乎会破坏(我)预期的行为。java.util.concurrent.DelayQueue忽略过期元素
道歉,该代码例如是如此长,但在总结:
- 主线程添加5个任务(AE)与各种延迟一个DelayQueue(0毫秒,10毫秒,100毫秒1000毫秒,10000ms)
- 另一胎开始它增加了后3000ms
- 另一个任务的DelayQueue主线程轮询DelayQueue并在每个任务报告到期
- 后8000MS主线程报告中剩余的DelayQueue 任务
,我从代码示例得到的输出是:
------initial tasks ---------------
task A due in 0ms
task B due in 9ms
task C due in 99ms
task D due in 999ms
task E due in 9999ms
task F due in 99999ms
------processing--------------------
time = 5 task A due in -1ms
time = 14 task B due in 0ms
time = 104 task C due in 0ms
time = 1004 task D due in 0ms
time = 3003 added task Z due in 0ms
------remaining after 15007ms -----------
task F due in 84996ms
task E due in -5003ms
task Z due in -12004ms
我的问题是:为什么后15000ms都存在过期留在DelayQueue任务(即其中GetDelay()返回一个值-ve) ?
,我查了一些事情:
- 我已经实现的compareTo()来定义任务的自然顺序
- equals()方法是用的compareTo()
- hashCode()方法是一贯的重写
我会对如何解决这个问题感兴趣。预先感谢您的帮助。 (和所有这些堆栈溢出的答案已经帮助我去约会了:)
package test;
import java.util.concurrent.DelayQueue;
import java.util.concurrent.Delayed;
import java.util.concurrent.TimeUnit;
public class Test10_DelayQueue {
private static final TimeUnit delayUnit = TimeUnit.MILLISECONDS;
private static final TimeUnit ripeUnit = TimeUnit.NANOSECONDS;
static long startTime;
static class Task implements Delayed {
public long ripe;
public String name;
public Task(String name, int delay) {
this.name = name;
ripe = System.nanoTime() + ripeUnit.convert(delay, delayUnit);
}
@Override
public boolean equals(Object obj) {
if (obj instanceof Task) {
return compareTo((Task) obj) == 0;
}
return false;
}
@Override
public int hashCode() {
int hash = 7;
hash = 67 * hash + (int) (this.ripe^(this.ripe >>> 32));
hash = 67 * hash + (this.name != null ? this.name.hashCode() : 0);
return hash;
}
@Override
public int compareTo(Delayed delayed) {
if (delayed instanceof Task) {
Task that = (Task) delayed;
return (int) (this.ripe - that.ripe);
}
throw new UnsupportedOperationException();
}
@Override
public long getDelay(TimeUnit unit) {
return unit.convert(ripe - System.nanoTime(), ripeUnit);
}
@Override
public String toString() {
return "task " + name + " due in " + String.valueOf(getDelay(delayUnit) + "ms");
}
}
static class TaskAdder implements Runnable {
DelayQueue dq;
int delay;
public TaskAdder(DelayQueue dq, int delay) {
this.dq = dq;
this.delay = delay;
}
@Override
public void run() {
try {
Thread.sleep(delay);
Task z = new Task("Z", 0);
dq.add(z);
Long elapsed = System.currentTimeMillis() - startTime;
System.out.println("time = " + elapsed + "\tadded " + z);
} catch (InterruptedException e) {
}
}
}
public static void main(String[] args) {
startTime = System.currentTimeMillis();
DelayQueue<Task> taskQ = new DelayQueue<Task>();
Thread thread = new Thread(new TaskAdder(taskQ, 3000));
thread.start();
taskQ.add(new Task("A", 0));
taskQ.add(new Task("B", 10));
taskQ.add(new Task("C", 100));
taskQ.add(new Task("D", 1000));
taskQ.add(new Task("E", 10000));
taskQ.add(new Task("F", 100000));
System.out.println("------initial tasks ---------------");
Task[] tasks = taskQ.toArray(new Task[0]);
for (int i = 0; i < tasks.length; i++) {
System.out.println(tasks[i]);
}
System.out.println("------processing--------------------");
try {
Long elapsed = System.currentTimeMillis() - startTime;
while (elapsed < 15000) {
Task task = taskQ.poll(1, TimeUnit.SECONDS);
elapsed = System.currentTimeMillis() - startTime;
if (task != null) {
System.out.println("time = " + elapsed + "\t" + task);
}
}
System.out.println("------remaining after " + elapsed + "ms -----------");
tasks = taskQ.toArray(new Task[0]);
for (int i = 0; i < tasks.length; i++) {
System.out.println(tasks[i]);
}
} catch (InterruptedException e) {
}
}
}
非常感谢 - 您的答案解决了我的问题。但我不明白为什么。 API声明CompareTo“返回一个负整数,零或正整数,因为该对象小于,等于或大于指定对象”。我理解重新使用Long.compareTo()的智慧,但我不明白为什么我的代码不符合compareTo契约!? – nhoj
原因是数值溢出。你正在将一个“长”的差异以纳秒为单位转换为“int”,但是在int中不能持有超过2.2秒的纳秒,并且会产生一个溢出 - 给出或多或少的随机结果,所以队列中的订单可能*后面*一个有一个以后到期。 poll()不会超出队列中的下一个项目,其顺序是在项目放入队列时定义的。 – Bohemian
溢出。将'long'投射到'int'可以改变它的符号。只有大约20亿个积极的'int',大约2秒的纳秒。你很可能有'this.ripe> that.ripe'但是'(int)(this.ripe - that.ripe)'为负的值。 –