jdk 1.8 hashmap get 方法

 * Returns the value to which the specified key is mapped,
 * or {@code null} if this map contains no mapping for the key.
 *
 * <p>More formally, if this map contains a mapping from a key
 * {@code k} to a value {@code v} such that {@code (key==null ? k==null :
 * key.equals(k))}, then this method returns {@code v}; otherwise
 * it returns {@code null}.  (There can be at most one such mapping.)
 *
 * <p>A return value of {@code null} does not <i>necessarily</i>
 * indicate that the map contains no mapping for the key; it's also
 * possible that the map explicitly maps the key to {@code null}.
 * The {@link #containsKey containsKey} operation may be used to
 * distinguish these two cases.
 *
 * @see #put(Object, Object)
 */
public V (Object key) {
    Node<K,V> e;
    // 将 key hash ，获取哈希数组桶，如果桶为空则直接返回 null
    return (e = getNode(hash(key), key)) == null ? null : e.value;
}


 * Implements Map.get and related methods
 *
 * @param hash hash for key
 * @param key the key
 * @return the node, or null if none
 */
final Node<K,V> getNode(int hash, Object key) {
    Node<K,V>[] tab; Node<K,V> first, e; int n; K k;

    if ((tab = table) != null && (n = tab.length) > 0 &&
        (first = tab[(n - 1) & hash]) != null) {

        // 判断桶的第一个位置是否为查询的 key ，是旧直接返回
        if (first.hash == hash && // always check first node
            ((k = first.key) == key || (key != null && key.equals(k))))
            return first;
        if ((e = first.next) != null) {

            // 判断下一个是红黑树还是链表
            if (first instanceof TreeNode)
                // 红黑树按照树的查找方式返回值
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);
            do {
                // 链表按照链表遍历的方式返回值
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}