Hashmap in java

    /**
     * The table, initialized on first use, and resized as
     * necessary. When allocated, length is always a power of two.
     * (We also tolerate length zero in some operations to allow
     * bootstrapping mechanics that are currently not needed.)
     */
    transient Node<K,V>[] table;　　//static class Node<K,V> implements Map.Entry<K,V>

 1  /**
 2      * The default initial capacity - MUST be a power of two.
 3      */
 4     static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16
 5 
 6     /**
 7      * The maximum capacity, used if a higher value is implicitly specified
 8      * by either of the constructors with arguments.
 9      * MUST be a power of two <= 1<<30.
10      */
11     static final int MAXIMUM_CAPACITY = 1 << 30;
12 
13     /**
14      * The bin count threshold for using a tree rather than list for a
15      * bin.  Bins are converted to trees when adding an element to a
16      * bin with at least this many nodes. The value must be greater
17      * than 2 and should be at least 8 to mesh with assumptions in
18      * tree removal about conversion back to plain bins upon
19      * shrinkage.
20      */
21     static final int TREEIFY_THRESHOLD = 8;　　//每个索引处链表元素个数的阈值，超过则用二叉查找树存

    /**
     * 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);
    }

 1  public V put(K key, V value) {
 2         return putVal(hash(key), key, value, false, true);
 3     }
 4 
 5     /**
 6      * Implements Map.put and related methods
 7      *
 8      * @param hash hash for key
 9      * @param key the key
10      * @param value the value to put
11      * @param onlyIfAbsent if true, don‘t change existing value
12      * @param evict if false, the table is in creation mode.
13      * @return previous value, or null if none
14      */
15     final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
16                    boolean evict) {
17         Node<K,V>[] tab; Node<K,V> p; int n, i;
　　　// 如果目前table为空或0，就对table初始化resize()，
18         if ((tab = table) == null || (n = tab.length) == 0)
19             n = (tab = resize()).length;
     //如果当前key值的hash索引处没有值，直接在该索引处存储键值对
20         if ((p = tab[i = (n - 1) & hash]) == null)
21             tab[i] = newNode(hash, key, value, null);
     //如果当前key值的hash索引处有值
22         else {
23             Node<K,V> e; K k;
     //如果当前索引处的key值和待插入的key值相同（即待插入键值对与bin内第一个键值对的键值相同）----即使是不同对象，值相同就可以（因为采用了equals()方法判断）
24             if (p.hash == hash &&
25                 ((k = p.key) == key || (key != null && key.equals(k))))
26                 e = p;
     //否则，如果当前bin内是以树节点的形式存储，把该节点插入到这个树中
27             else if (p instanceof TreeNode)
28                 e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);　　//如果树中已经有这个节点，返回这个节点之前的value；如果没有返回null
　　　//否则，即当前bin内以链表形式存储，遍历链表，插入该节点
29             else {
30                 for (int binCount = 0; ; ++binCount) {
31                     if ((e = p.next) == null) {
32                         p.next = newNode(hash, key, value, null);
33                         if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st　　//如果插入节点之后，元素个数超过阈值，就把链表转换为二叉查找树
34                             treeifyBin(tab, hash);
35                         break;
36                     }
37                     if (e.hash == hash &&
38                         ((k = e.key) == key || (key != null && key.equals(k))))
39                         break;
40                     p = e;
41                 }
42             }
　　　　//如果该索引处有键值相同的键值对，那么更新原有键值对的value为新的value，返回旧的value
43             if (e != null) { // existing mapping for key
44                 V oldValue = e.value;
45                 if (!onlyIfAbsent || oldValue == null)
46                     e.value = value;
47                 afterNodeAccess(e);
48                 return oldValue;
49             }
50         }
51         ++modCount;
52         if (++size > threshold)
53             resize();
54         afterNodeInsertion(evict);
55         return null;
56     }

 1     /**
 2      * Returns the value to which the specified key is mapped,
 3      * or {@code null} if this map contains no mapping for the key.
 4      *
 5      * <p>More formally, if this map contains a mapping from a key
 6      * {@code k} to a value {@code v} such that {@code (key==null ? k==null :
 7      * key.equals(k))}, then this method returns {@code v}; otherwise
 8      * it returns {@code null}.  (There can be at most one such mapping.)
 9      *
10      * <p>A return value of {@code null} does not <i>necessarily</i>
11      * indicate that the map contains no mapping for the key; it‘s also
12      * possible that the map explicitly maps the key to {@code null}.
13      * The {@link #containsKey containsKey} operation may be used to
14      * distinguish these two cases.
15      *
16      * @see #put(Object, Object)
17      */
18     public V get(Object key) {
19         Node<K,V> e;
20         return (e = getNode(hash(key), key)) == null ? null : e.value;
21     }