【Java】Java Queue的简介

阻塞队列

阻塞队列有几个实现:

  • ArrayBlockingQueue
  • LinkedBlockingQueue
  • PriorityBlockingQueue
  • DelayQueue
  • SynchronousQueue
  • LinkedTransferQueue
  • LinkedBlockingDeque

他们的共同父类是AbstractQueue。我们一起看ArrayBlockingQueue的实现。

ArrayBlockingQueue,数组、有界、出入队阻塞

数据存储

数据存储在数组中,用几个变量标记下一个获取或存储的元素:

    /** The queued items */
    final Object[] items; // 用数组存储元素

    /** items index for next take, poll, peek or remove */
    int takeIndex; // 返回元素的下标

    /** items index for next put, offer, or add */
    int putIndex; // 插入元素的下标

    /** Number of elements in the queue */
    int count; // 数量

阻塞逻辑

添加、删除元素需要使用ReentrantLock加锁,满队列、空队列情况的等待与唤醒使用各自的Condition:

    public ArrayBlockingQueue(int capacity, boolean fair) {
        ...
        lock = new ReentrantLock(fair);
        notEmpty = lock.newCondition();
        notFull =  lock.newCondition();
    }

插入元素,返回是否成功

    public boolean offer(E e) {
        checkNotNull(e);
        final ReentrantLock lock = this.lock;
        lock.lock();
        try {
            if (count == items.length) // 队列满,插入不成,返回false
                return false;
            else {
                enqueue(e);
                return true; // 插入成功,返回true
            }
        } finally {
            lock.unlock();
        }
    }

插入元素,成功返回true,失败抛出异常

它调用offer方法,插入成功返回true,失败抛出异常:

    public boolean add(E e) {
        if (offer(e))
            return true;
        else
            throw new IllegalStateException("Queue full");
    }

插入元素,队列满了则阻塞

    public void put(E e) throws InterruptedException {
        checkNotNull(e); // e为空抛出异常
        final ReentrantLock lock = this.lock;
        lock.lockInterruptibly();
        try {
            while (count == items.length) // 当队列满了
                notFull.await(); // notFull的Condition等待条件成熟
            enqueue(e); // 条件成熟了才插入元素
        } finally {
            lock.unlock();
        }
    }

插入元素,队列满了则阻塞指定超时时间

主体逻辑与put(E e)一致,只是加了超时逻辑:

    public boolean offer(E e, long timeout, TimeUnit unit)
        throws InterruptedException {

        checkNotNull(e);
        long nanos = unit.toNanos(timeout); // 将超时时间转换为Nano单位
        final ReentrantLock lock = this.lock;
        lock.lockInterruptibly();
        try {
            while (count == items.length) {
                if (nanos <= 0) // 超时了,返回false
                    return false;
                nanos = notFull.awaitNanos(nanos); // Condition等待指定时间
            }
            enqueue(e);
            return true; // 超时时间内插入成功,返回true
        } finally {
            lock.unlock();
        }
    }

删除元素,返回是否删除成功

    public boolean remove(Object o) {
        if (o == null) return false;
        final Object[] items = this.items;
        final ReentrantLock lock = this.lock;
        lock.lock();
        try {
            if (count > 0) {
                final int putIndex = this.putIndex;
                int i = takeIndex;
                do {
                    if (o.equals(items[i])) { // 遍历到要删除的元素,删除并返回true
                        removeAt(i);
                        return true;
                    }
                    if (++i == items.length)
                        i = 0;
                } while (i != putIndex);
            }
            return false; // 遍历完毕,没有找到,返回false
        } finally {
            lock.unlock();
        }
    }

删除元素,返回删除的元素,没匹配到返回null

    public E poll() {
        final ReentrantLock lock = this.lock;
        lock.lock();
        try {
            return (count == 0) ? null : dequeue();
        } finally {
            lock.unlock();
        }
    }

删除元素,队列为空则阻塞

    public E take() throws InterruptedException {
        final ReentrantLock lock = this.lock;
        lock.lockInterruptibly();
        try {
            while (count == 0) // 队列为空
                notEmpty.await(); // 元素出队,队列元素为空,等待非空条件成熟
            return dequeue();
        } finally {
            lock.unlock();
        }
    }

入队逻辑,入队成功后同时非空条件成熟:

    private void enqueue(E x) { // 入队
        // assert lock.getHoldCount() == 1;
        // assert items[putIndex] == null;
        final Object[] items = this.items;
        items[putIndex] = x;
        if (++putIndex == items.length)
            putIndex = 0;
        count++;
        notEmpty.signal(); // 元素入队后,通知非空条件已成熟
    }

删除元素,队列为空阻塞指定超时时间

主体逻辑与take()一直,但有等待超时逻辑:

    public E poll(long timeout, TimeUnit unit) throws InterruptedException {
        long nanos = unit.toNanos(timeout); // 转化为nano单位
        final ReentrantLock lock = this.lock;
        lock.lockInterruptibly();
        try {
            while (count == 0) {
                if (nanos <= 0)
                    return null;
                nanos = notEmpty.awaitNanos(nanos); // 等待指定超时时间
            }
            return dequeue();
        } finally {
            lock.unlock();
        }
    }

LinkedBlockingQueue,链表、有界、出入队阻塞

存储结构

用链表作为存储,Node是链表的节点:

    static class Node<E> {
        E item; // 元素值

        /**
         * One of:
         * - the real successor Node
         * - this Node, meaning the successor is head.next
         * - null, meaning there is no successor (this is the last node)
         */
        Node<E> next; // 下一节点

        Node(E x) { item = x; } // 构造方法
    }

PriorityBlockingQueue,无界,出队阻塞

出队阻塞

它是无界的,所以只有出队时队列无元素才会堵塞,依赖notEmpty的Condition:

    /**
     * Condition for blocking when empty
     */
    private final Condition notEmpty;

优先级顺序

它的优先级依赖比较器:

    /**
     * The comparator, or null if priority queue uses elements'
     * natural ordering.
     */
    private transient Comparator<? super E> comparator;

DelayQueue,无界、出队阻塞、等待指定时间才能出队

数据存储

它的数据实现依赖于PriorityQueue,所以队列的元素需实现Comparable:

    private final PriorityQueue<E> q = new PriorityQueue<E>();

出队

    public E poll() {
        final ReentrantLock lock = this.lock;
        lock.lock();
        try {
            E first = q.peek(); // 获取下一个即将的出队元素
            if (first == null || first.getDelay(NANOSECONDS) > 0) // 如果无出队元素,或出队元素的时间未到
                return null;
            else
                return q.poll(); // 实际的出队
        } finally {
            lock.unlock();
        }
    }

阻塞出队

    public E take() throws InterruptedException {
        final ReentrantLock lock = this.lock;
        lock.lockInterruptibly();
        try {
            for (;;) {
                E first = q.peek(); // 获取将要出队的元素
                if (first == null) // 为空,则等待
                    available.await();
                else {
                    long delay = first.getDelay(NANOSECONDS);
                    if (delay <= 0) // 时间已到,出队,跳出方法
                        return q.poll();
                    first = null; // don't retain ref while waiting // 等待期间取消引用
                    if (leader != null) # TODO,未理解透彻
                        available.await();
                    else {
                        Thread thisThread = Thread.currentThread();
                        leader = thisThread; // 当前线程赋予leader
                        try {
                            available.awaitNanos(delay); // 等待剩余时间
                        } finally {
                            if (leader == thisThread)
                                leader = null;
                        }
                    }
                }
            }
        } finally {
            if (leader == null && q.peek() != null)
                available.signal();
            lock.unlock();
        }
    }

SynchronousQueue,阻塞队列,不存储元素

依赖于TransferQueue和TransferStack

它可设置是否公平,分别依赖于TransferQueue和TransferStack,默认非公平

    public SynchronousQueue(boolean fair) {
        transferer = fair ? new TransferQueue<E>() : new TransferStack<E>();
    }

阻塞入队和出队

    public void put(E e) throws InterruptedException {
        if (e == null) throw new NullPointerException();
        if (transferer.transfer(e, false, 0) == null) {
            Thread.interrupted();
            throw new InterruptedException();
        }
    }
    public E take() throws InterruptedException {
        E e = transferer.transfer(null, false, 0);
        if (e != null)
            return e;
        Thread.interrupted();
        throw new InterruptedException();
    }
posted @ 2017-08-19 00:05  nick_huang  阅读(411)  评论(0编辑  收藏  举报