剖析HashMap底层原理

HashMap

java中使用非常之频繁的一种数据接口，面试中也频繁考察的一种数据结构。处于好奇的心态，简单分析一下它的源码。

JDK1.8

本篇博客的基于JDK 1.8版本，其他版本暂时不做分析。

开始

日常使用HashMap，用得比较多的方法就是put/get了，那就从put开始入手：

    /**
     * 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);
    }

根据注释还是比较好理解的，解释了一下key会与value做关联，如果key已经存在，那么新的value会覆盖掉旧的value，并且返回出来旧的value。另外解释了一下value可以是null，也会被返回出来。再去看一下putVal中的hash(key)方法

     /**
     * Computes key.hashCode() and spreads (XORs) higher bits of hash
     * to lower.  Because the table uses power-of-two masking, sets of
     * hashes that vary only in bits above the current mask will
     * always collide. (Among known examples are sets of Float keys
     * holding consecutive whole numbers in small tables.)  So we
     * apply a transform that spreads the impact of higher bits
     * downward. There is a tradeoff between speed, utility, and
     * quality of bit-spreading. Because many common sets of hashes
     * are already reasonably distributed (so don't benefit from
     * spreading), and because we use trees to handle large sets of
     * collisions in bins, we just XOR some shifted bits in the
     * cheapest possible way to reduce systematic lossage, as well as
     * to incorporate impact of the highest bits that would otherwise
     * never be used in index calculations because of table bounds.
     */
    static final int hash(Object key) {
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

注释中讲得也很明白，主要是避免频繁在一些场景下出现key计算出的hash值仅仅高位变动，低位相同的情况造成的严重碰撞。于是直接把hash值的无符号右移16位(高位0补齐)与原hash值异或，扩大高16位的影响力得出最终的hash值。下面进入putVal()方法:

    /**
     * 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<K,V>[] tab; Node<K,V> p; int n, i;
        // 这边做了一个判断，可以看出HashMap桶空间的初始化是在put时进行的
        // 也就是实例化时并没有创建内部空间，用时再创建，节省空间
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        // 下面就是寻址了，(n-1) & hash非常巧妙，也解释了为什么HashMap内部桶个数是2的幂次
        // n为2的幂次减去1后的二进制数就都是1了
        // 举个例子16的二进制数是10000，减一后为1111(高位补0)，hash 32位
        // 1111 & hash 后必然是前28位都是0，后4位为真正结果，但是也必然是0-15中的一个数
        // 也就对应了16个桶中的一个下标了，非常巧妙
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
        // 如果出现碰撞则形成链表，链表超过了8位则转红黑树
        // 至于为何是8位其实源码注释里也说明白了
            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;
    }