HashMap源码

hash(Object key) 

int类型4个字节，32位，将h右移动16位。再异或运算。即 高16位与低16位进行异或，每一位都充分参与运算，这样得到的值更加散列。

    static final int hash(Object key) {
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

putVal()方法

下列代码行4、5：

刚进来时table为null，调用resize()方法中返回newTab，则n为16。

DEFAULT_INITIAL_CAPACITY=16

newCap = DEFAULT_INITIAL_CAPACITY;

Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];

代码行6：&运算，即取%。且效率比%高。

用于判断当前tab[i]是否为null

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);//创建节点，引出HashMap是一个数组+链表的结构
        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)//红黑树节点
                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);
                        //循环多少次---->链表上有多少个节点
                        if (binCount >= TREEIFY_THRESHOLD - 1) 
                            treeifyBin(tab, hash);//当到达阈值8时，链表转化成红黑树
                        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)//onlyIfAbsent:是否替换旧值，默认替换
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)//threshold=12*0.75=12
            resize();
        afterNodeInsertion(evict);
        return null;
    }

resize()方法

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) {
            //边界值的判断
            if (oldCap >= MAXIMUM_CAPACITY) {
                threshold = Integer.MAX_VALUE;
                return oldTab;
            }
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY && //newCap=2*16
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                newThr = oldThr << 1; // 对应阈值也发生变化=2*12
        }
        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);
        }
        if (newThr == 0) {//一些边界值的判断
            float ft = (float)newCap * loadFactor;
            newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
                      (int)ft : Integer.MAX_VALUE);
        }
        threshold = newThr;
        @SuppressWarnings({"rawtypes","unchecked"})
            Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];//创建一个新的容量大小的数组
        table = newTab;
        //将之前的数组内容重新放置在新数组。
        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) //红黑树则进行分割
                        ((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;//新的头节点，尾节点
                        Node<K,V> next;
                        do {
                            next = e.next;
                            //妙笔
                            if ((e.hash & oldCap) == 0) {//还是放在原来的位置
                                if (loTail == null)
                                    loHead = e;
                                else
                                    loTail.next = e;
                                loTail = e;
                            }
                            else {//放在新的扩容的部分
                                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;
    }

上图代码中：(e.hash & oldCap) == 0 用于判断当前节点是否需要重新被放置到新扩容的部分 

假设之前有节点 hash1=6，hash2=22，oldCap=16

未扩容前，处于同一个下标

结果为0，说明还是在原来的部分，否则放置 newTab[j + oldCap] = hiHead;