Java集合中，isEmpty()与size()==0的区别

小半 • 2023年1月31日下午5:07 • 后端开发 • 阅读 110

关于集合的处理，Java开发手册有这么一段话:

【强制】判断所有集合内部的元素是否为空，使用 isEmpty()方法，而不是 size()==0 的方式。

说明：在某些集合中，前者的时间复杂度为 O(1)，而且可读性更好。

下面我们通过一些源码来看看

HashMap源码

    /**
     * Returns the number of key-value mappings in this map.
     *
     * @return the number of key-value mappings in this map
     */
    public int size() {
        return size;
    }

    /**
     * Returns <tt>true</tt> if this map contains no key-value mappings.
     *
     * @return <tt>true</tt> if this map contains no key-value mappings
     */
    public boolean isEmpty() {
        return size == 0;
    }

ConcurrentHashMap源码


    /**
     * {@inheritDoc}
     */
    public int size() {
        long n = sumCount();
        return ((n < 0L) ? 0 :
                (n > (long)Integer.MAX_VALUE) ? Integer.MAX_VALUE :
                (int)n);
    }

    /**
     * {@inheritDoc}
     */
    public boolean isEmpty() {
        return sumCount() <= 0L; // ignore transient negative values
    }

ConcurrentLinkedQueue源码

    /**
     * Returns {@code true} if this queue contains no elements.
     *
     * @return {@code true} if this queue contains no elements
     */
    public boolean isEmpty() {
        return first() == null;
    }

    /**
     * Returns the number of elements in this queue.  If this queue
     * contains more than {@code Integer.MAX_VALUE} elements, returns
     * {@code Integer.MAX_VALUE}.
     *
     * <p>Beware that, unlike in most collections, this method is
     * <em>NOT</em> a constant-time operation. Because of the
     * asynchronous nature of these queues, determining the current
     * number of elements requires an O(n) traversal.
     * Additionally, if elements are added or removed during execution
     * of this method, the returned result may be inaccurate.  Thus,
     * this method is typically not very useful in concurrent
     * applications.
     *
     * @return the number of elements in this queue
     */
    public int size() {
        int count = 0;
        for (Node<E> p = first(); p != null; p = succ(p))
            if (p.item != null)
                // Collection.size() spec says to max out
                if (++count == Integer.MAX_VALUE)
                    break;
        return count;
    }