HashSet中add()方法的使用

HashSet 傳入對象,怎麼過濾

首先,我們知道所有的Set裏面都套了一個Map,使用Map的Key存值,Value存放一個固定的Object,這是適配器模式。源碼如下：

public class HashSet<E>
    extends AbstractSet<E>
    implements Set<E>, Cloneable, java.io.Serializable
{
    static final long serialVersionUID = -5024744406713321676L;

    private transient HashMap<E,Object> map;

    // Dummy value to associate with an Object in the backing Map
    private static final Object PRESENT = new Object();

    /**
     * Constructs a new, empty set; the backing <tt>HashMap</tt> instance has
     * default initial capacity (16) and load factor (0.75).
     */
    public HashSet() {
        map = new HashMap<>();
    }

    /**
     * Constructs a new set containing the elements in the specified
     * collection.  The <tt>HashMap</tt> is created with default load factor
     * (0.75) and an initial capacity sufficient to contain the elements in
     * the specified collection.
     *
     * @param c the collection whose elements are to be placed into this set
     * @throws NullPointerException if the specified collection is null
     */
    public HashSet(Collection<? extends E> c) {
        map = new HashMap<>(Math.max((int) (c.size()/.75f) + 1, 16));
        addAll(c);
    }

    /**
     * Constructs a new, empty set; the backing <tt>HashMap</tt> instance has
     * the specified initial capacity and the specified load factor.
     *
     * @param      initialCapacity   the initial capacity of the hash map
     * @param      loadFactor        the load factor of the hash map
     * @throws     IllegalArgumentException if the initial capacity is less
     *             than zero, or if the load factor is nonpositive
     */
    public HashSet(int initialCapacity, float loadFactor) {
        map = new HashMap<>(initialCapacity, loadFactor);
    }

    /**
     * Constructs a new, empty set; the backing <tt>HashMap</tt> instance has
     * the specified initial capacity and default load factor (0.75).
     *
     * @param      initialCapacity   the initial capacity of the hash table
     * @throws     IllegalArgumentException if the initial capacity is less
     *             than zero
     */
    public HashSet(int initialCapacity) {
        map = new HashMap<>(initialCapacity);
    }

    /**
     * Constructs a new, empty linked hash set.  (This package private
     * constructor is only used by LinkedHashSet.) The backing
     * HashMap instance is a LinkedHashMap with the specified initial
     * capacity and the specified load factor.
     *
     * @param      initialCapacity   the initial capacity of the hash map
     * @param      loadFactor        the load factor of the hash map
     * @param      dummy             ignored (distinguishes this
     *             constructor from other int, float constructor.)
     * @throws     IllegalArgumentException if the initial capacity is less
     *             than zero, or if the load factor is nonpositive
     */
    HashSet(int initialCapacity, float loadFactor, boolean dummy) {
        map = new LinkedHashMap<>(initialCapacity, loadFactor);
    }


......
}

那麼問題來了,如果傳入兩個對象,是什麼決定它們是不是相等的,舊的會不會替換新的對象,以及怎麼替換或者不替換呢?

我們來看下add方法

private static final Object PRESENT = new Object();

public boolean add(E e) {
    return map.put(e, PRESENT)==null;
}

實際調用的是HashMap的添加方法，只不過是value是成爲一個全局變量PRESENT。這時map是一個全局變量指向創建HashSet對象時創建的HashMap對象。

另外add方法的返回值類型是boolean，所以如果返回值是null那麼表示添加成功。

探究.add方法也即是探究.put方法。

    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> 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<K,V> 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<K,V>)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;
    }

下面我們結合一個例子分析源碼。

HashSet<String> set = new HashSet<>();
set.add("Tom");
set.add("Tom");
set.add("new String Tom");

• 第一次添加元素的過程:set.add(“Tom”);

首先定義Node<K,V>[] tab；意思是創建一個名爲tab的Node節點集合，然後if ((tab = table) == null || (n = tab.length) == 0)對比tab和table是否爲空，因爲table是全局變量所以程序運行開始之前初始值null，判斷爲true，所以不用判斷(n = tab.length) == 0，直接執行 n = (tab = resize()).length;，resize方法打開底層並沒有看懂，但是知道是返回值是newTab，其實此時tab已經是被替換了成爲長度爲16的數組，n等於tab的長度最大值是16，接下來判斷p = tab[i = (n - 1) & hash]) == null 其中因爲i的範圍是0~15,所以n-1來控制防止數組溢出，至於&hash是程序自動編譯計算得來的tab的地址(在第三種情況我會詳細描述)。這樣就執行tab[i] = newNode(hash, key, value, null);不執行else，直接執行圖中代碼（源代碼的最後一段）
返回值是null，所以put方法完成返回值null，然後boolean判斷是turn，.add方法完成。