HashMap 源码解毒

PUT 方法解毒：

hashcode 高低16进行异或运算，尽量降低哈希冲突的概率

如果数组很小，hashcode的高位就不能被很好利用。

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        // 如果tab为空或者长度为0，则需要初始化（默认大小16，默认扩容门槛0.75*16）
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        // 通过（(n - 1) & hash）找到数组中合适的位子
        if ((p = tab[i = (n - 1) & hash]) == null)
            // 如果数组中为空（数组中会存放Node链表或者红黑树） 新建Node节点
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            // 如果第一个节点的值等于插入的value
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            // 如果节点是树节点，红黑树插入
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    // 如果走到链表末尾，仍然发现没有相同值，则创建节点插入
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        // 达到树化阈值进行树化
                        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;
                }
            }
            // 如果原来存在Node，是不是需要替换？onlyIfAbsent
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        // 大于阈值进行扩容  比如阈值是12 size==12，而不是modCount==12，时候扩容
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

final Node<K,V>[] resize() {
        Node<K,V>[] oldTab = table;
        int oldCap = (oldTab == null) ? 0 : oldTab.length;
        int oldThr = threshold;
        int newCap, newThr = 0;
        if (oldCap > 0) {
            // 如果旧容量大于等于最大容量，就不能再扩容了
            if (oldCap >= MAXIMUM_CAPACITY) {
                threshold = Integer.MAX_VALUE;
                return oldTab;
            }
            // 容量是2倍旧容量
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                // 新阈值是旧阈值2倍
                newThr = oldThr << 1; // double threshold
        }
        else if (oldThr > 0) // initial capacity was placed in threshold
            newCap = oldThr;
        else {               // zero initial threshold signifies using defaults
            newCap = DEFAULT_INITIAL_CAPACITY;
            newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
        }
        // 计算新阈值
        if (newThr == 0) {
            float ft = (float)newCap * loadFactor;
            newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
                      (int)ft : Integer.MAX_VALUE);
        }
        threshold = newThr;
        @SuppressWarnings({"rawtypes","unchecked"})
        // 创建新数组
        Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
        table = newTab;
        if (oldTab != null) {
            for (int j = 0; j < oldCap; ++j) {
                Node<K,V> e;
                if ((e = oldTab[j]) != null) {
                    oldTab[j] = null;
                    if (e.next == null)
                        newTab[e.hash & (newCap - 1)] = e;
                    else if (e instanceof TreeNode)
                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                    else { // preserve order
                        Node<K,V> loHead = null, loTail = null;
                        Node<K,V> hiHead = null, hiTail = null;
                        Node<K,V> next;
                        do {
                            next = e.next;
                            if ((e.hash & oldCap) == 0) {
                                if (loTail == null)
                                    loHead = e;
                                else
                                    loTail.next = e;
                                loTail = e;
                            }
                            else {
                                if (hiTail == null)
                                    hiHead = e;
                                else
                                    hiTail.next = e;
                                hiTail = e;
                            }
                        } while ((e = next) != null);
                        if (loTail != null) {
                            loTail.next = null;
                            newTab[j] = loHead;
                        }
                        if (hiTail != null) {
                            hiTail.next = null;
                            newTab[j + oldCap] = hiHead;
                        }
                    }
                }
            }
        }
        return newTab;
    }