HashSet add() 方法底层调用HashMap时，存入的value为什么不是null?

看jdk源码，看到 LinkedHashSet时，说底层调用的LinkedHashMap进行存放数据，点入LinkedHashSet的add方法看了一下，看到如下代码：

    /**
     * Adds the specified element to this set if it is not already present.
     * More formally, adds the specified element e to this set if
     * this set contains no element e2 such that
     * (e==null ? e2==null : e.equals(e2)).
     * If this set already contains the element, the call leaves the set
     * unchanged and returns false.
     *
     * @param e element to be added to this set
     * @return true if this set did not already contain the specified
     * element
     */
    public boolean add(E e) {
        return map.put(e, PRESENT)==null;
    }

其中的PRESENT 为类的一个私有Object对象。

private static final Object PRESENT = new Object();

这就很值得研究一下，为什么add()方法中，存入的value不是null？正常来说，存入null的效率和空间占用是否会更好一点？

继续往下看，看map.put（）方法的实现：

    /**
     * Associates the specified value with the specified key in this map.
     * If the map previously contained a mapping for the key, the old
     * value is replaced.
     *
     * @param key key with which the specified value is to be associated
     * @param value value to be associated with the specified key
     * @return the previous value associated with key, or
     *         null if there was no mapping for key.
     *         (A null return can also indicate that the map
     *         previously associated null with key.)
     */
    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

putval()方法如下：

    /**
     * Implements Map.put and related methods
     *
     * @param hash hash for key
     * @param key the key
     * @param value the value to put
     * @param onlyIfAbsent if true, don't change existing value
     * @param evict if false, the table is in creation mode.
     * @return previous value, or null if none
     */
    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node[] tab; Node p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            else if (p instanceof TreeNode)
                e = ((TreeNode)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

实现不需要仔细看，先看他的关于return的注释：

     * @return previous value, or null if none

意思就是，返回当前值，如果值不存在，为第一次添加该key的话，返回null。

所以，如果前边add时，添加的value为null，那返回的话，就不能区分时第一次添加返回的空值null，还是重复添加时，返回的已有值，只是这个值为null，但是，添加的如果是一个Object，就可以避免这个问题了。

源码的每一步实现，都是有道理的。