延迟队列延迟后轮询

Posted 2023-02-24

【中文标题】延迟队列延迟后轮询

【英文标题】：Polling after a delay in delayed queue

【发布时间】：2014-08-04 06:11:35

【问题描述】：

我有一个示例程序，我试图从中了解延迟队列。我以特定的延迟向队列提供人员对象，当我尝试在 5 秒的间隔后轮询对象时，我应该得到延迟已过期的所有对象。但相反，我得到 null ，我不明白原因。但是当我将延迟设置为 0 时，此轮询有效。有人可以帮我弄清楚我在下面的示例代码中哪里出错了吗？

public class DelayedQueue 
    public static void main(String[] args) 
        BlockingQueue<Person> queue = new DelayQueue<Person>();
        Person a = new Person("ram", "chennai", 1);
        Person b = new Person("nick", "manali", 1);
        Person c = new Person("sam", "delhi", 2);
        try 
            queue.offer(a);
            queue.offer(b);
            queue.offer(c);
            System.out.println(queue.poll(5, TimeUnit.SECONDS));
         catch (InterruptedException e1) 
            // TODO Auto-generated catch block
            e1.printStackTrace();
        
    


class Person implements Delayed 
    private String name;
    private String place;
    private int runningTime;

    public Person(String name, String place, int runningTime) 
        this.name = name;
        this.place = place;
        this.runningTime = runningTime;
    
    public long getDelay(TimeUnit timeUnit)            
        return timeUnit.convert(this.runningTime, TimeUnit.MILLISECONDS);

    @Override
    public int compareTo(Delayed person) 
        Person b = (Person)person;   
        return this.name.compareTo(b.name);
    

    @Override
    public long getDelay(TimeUnit timeUnit)            
        return timeUnit.convert(this.runningTime, TimeUnit.MILLISECONDS);
    

【参考方案1】：

你的 getDelay 实现是错误的

@Override
    public long getDelay(TimeUnit timeUnit)        
        **// This will never return zero! and the element is never available.**
        return timeUnit.convert(this.runningTime, TimeUnit.MILLISECONDS);
    

尝试做这样的事情

@Override
    public long getDelay(TimeUnit timeUnit)            
        return timeUnit.convert(endOfDelay - System.currentTimeMillis(),
                          TimeUnit.MILLISECONDS);
    

endOfDelay 设置为 long ( System.currentTimeMillis() + delay (in ms)

这是您的代码的工作部分：

public class DelayedQueue

    public static void main(String[] args)
    
        BlockingQueue<Person> queue = new DelayQueue<Person>();
        Person a = new Person("ram", "chennai", 1);
        Person b = new Person("nick", "manali", 1);
        Person c = new Person("sam", "delhi", 2);
        try
        
            queue.offer(a);
            queue.offer(b);
            queue.offer(c);
            System.out.println(queue.poll(2, TimeUnit.SECONDS));
         catch (InterruptedException e1)
        
            // TODO Auto-generated catch block
            e1.printStackTrace();
        
    

class Person implements Delayed

    private String name;
    private String place;
    private long delayTime;

    public Person(String name, String place, long delayTime)
    
        this.name = name;
        this.place = place;
        this.delayTime = System.currentTimeMillis() + TimeUnit.SECONDS.toMillis(delayTime);
    

    @Override
    public int compareTo(Delayed person)
    
        Person b = (Person) person;
        return this.name.compareTo(b.name);
    

    @Override
    public long getDelay(TimeUnit timeUnit)
    
        return timeUnit.convert(delayTime - System.currentTimeMillis(), TimeUnit.MILLISECONDS);
    

【讨论】：

我在 javadocs 中看到“队列的头部是延迟过期的元素”。但我没有得到使用轮询检索到的最少过期元素。因此，根据上面的程序，当我执行 queue.poll(2, TimeUnit.SECONDS) 时，我无法获得元素 ram 但 nick 并且当我更改延迟时，我总是得到 nick。你能解释一下为什么 poll() 的这种行为是随机的吗？

这是因为您定义了 compareTo。因为它尼克是峰值元素。因此，如果延迟