5 分鐘掌握 JDK 源碼之 ArrayBlockingQueue

ArrayBlockingQueue

初始化

1. 有界隊列，隊列容量一旦初始化不能再改變
2. 支持對鎖的公平性進行設置

容量擴展

無

實現原理

1. 用一個固定長度的數組保存隊列元素
2. 用 putIndex 和 takeIndex 分別標記隊列頭和尾

// 用於標記隊列尾在數組的索引
int takeIndex;

// 用於標記隊列頭在數組的索引
int putIndex;

// 數組元素個數
int count

// 鎖
final ReentrantLock lock;

// 當隊裏爲空的時候，等待該條件
private final Condition notEmpty;

// 當隊裏滿的時候，等待該條件
private final Condition notFull;

入隊列

實現原理非常簡單。分析略

算法複雜度：O(1)

   private void enqueue(E x) {
        // 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();
    }

    public boolean offer(E e) {
        checkNotNull(e);
        final ReentrantLock lock = this.lock;
        lock.lock();
        try {
            if (count == items.length)
                return false;
            else {
                enqueue(e);
                return true;
            }
        } finally {
            lock.unlock();
        }
    }

    public void put(E e) throws InterruptedException {
        checkNotNull(e);
        final ReentrantLock lock = this.lock;
        lock.lockInterruptibly();
        try {
            while (count == items.length)
                notFull.await();
            enqueue(e);
        } finally {
            lock.unlock();
        }
    }

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

        checkNotNull(e);
        long nanos = unit.toNanos(timeout);
        final ReentrantLock lock = this.lock;
        lock.lockInterruptibly();
        try {
            while (count == items.length) {
                if (nanos <= 0)
                    return false;
                nanos = notFull.awaitNanos(nanos);
            }
            enqueue(e);
            return true;
        } finally {
            lock.unlock();
        }
    }

出隊列

實現原理非常簡單。分析略

算法複雜度：O(1)

    private E dequeue() {
        // assert lock.getHoldCount() == 1;
        // assert items[takeIndex] != null;
        final Object[] items = this.items;
        @SuppressWarnings("unchecked")
        E x = (E) items[takeIndex];
        items[takeIndex] = null;
        if (++takeIndex == items.length)
            takeIndex = 0;
        count--;
        if (itrs != null)
            itrs.elementDequeued();
        notFull.signal();
        return x;
    }

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

    public E take() throws InterruptedException {
        final ReentrantLock lock = this.lock;
        lock.lockInterruptibly();
        try {
            while (count == 0)
                notEmpty.await();
            return dequeue();
        } finally {
            lock.unlock();
        }
    }

    public E poll(long timeout, TimeUnit unit) throws InterruptedException {
        long nanos = unit.toNanos(timeout);
        final ReentrantLock lock = this.lock;
        lock.lockInterruptibly();
        try {
            while (count == 0) {
                if (nanos <= 0)
                    return null;
                nanos = notEmpty.awaitNanos(nanos);
            }
            return dequeue();
        } finally {
            lock.unlock();
        }
    }

remove

刪除操作需要注意的是，可能會進行大量的元素移動

算法複雜度：O(N)

   void removeAt(final int removeIndex) {
        // assert lock.getHoldCount() == 1;
        // assert items[removeIndex] != null;
        // assert removeIndex >= 0 && removeIndex < items.length;
        final Object[] items = this.items;
        if (removeIndex == takeIndex) {
            // removing front item; just advance
            items[takeIndex] = null;
            if (++takeIndex == items.length)
                takeIndex = 0;
            count--;
            if (itrs != null)
                itrs.elementDequeued();
        } else {
            // an "interior" remove

            // slide over all others up through putIndex.
            final int putIndex = this.putIndex;
            for (int i = removeIndex;;) {
                int next = i + 1;
                if (next == items.length)
                    next = 0;
                if (next != putIndex) {
                    items[i] = items[next];
                    i = next;
                } else {
                    items[i] = null;
                    this.putIndex = i;
                    break;
                }
            }
            count--;
            if (itrs != null)
                itrs.removedAt(removeIndex);
        }
        notFull.signal();
    }
    
    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])) {
                        removeAt(i);
                        return true;
                    }
                    if (++i == items.length)
                        i = 0;
                } while (i != putIndex);
            }
            return false;
        } finally {
            lock.unlock();
        }
    }

總結

基於數組的有界的線程安全的隊列，實現非常簡單，有如下要點需要注意

1. 隊列容量一旦確定，不可再修改。
2. 非常適合生產者和消費者模式。
3. 支持對鎖的公平性進行設置。