我把纯源码放到了随笔： https://www.cnblogs.com/zhangxuezhi/p/11660818.html
首先HashMap 继承了一个抽象类：AbstractMap，实现了3个接口：Map，Cloneable，Serializable。因此hashmap本身是可克隆、可序列化的，并且继承了AbstractMap的非私有方法和非私有域，然后需要去实现map接口规定的方法，
先来看put函数的源码：
/**
* 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 <tt>key</tt>, or
* <tt>null</tt> if there was no mapping for <tt>key</tt>.
* (A <tt>null</tt> return can also indicate that the map
* previously associated <tt>null</tt> with <tt>key</tt>.)
*/
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
该函数做的事情有两件，首先是计算key的hash值，然后调用putVal函数，先来看看hash计算的函数：
/**
* 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);
}
如果key为null，返回0；否则，取key的hashCode，高16位不变，然后将其高16位和低16位进行异或运算，用运算后的值替代原始值的低16位。（>>>为无符号右移，即右移时，在左边填充0。所以hashCode值右移16位之后，其左边16位为0，而与0进行异或运算，值不变。）这样做其实低16位值就包含了高16位值的信息，因此在容器的大小不超过2^16的情况下，其hash值就同时包含高低位的信息量，该值更能充分利用hashCode的整体信息，从一定程度上减少hash碰撞（个人理解）。
接下来看putVal函数
/**
* 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;
if ((tab = table) == null || (n = tab.length) == 0)
//table 为空
//n被赋值为table的长度
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null)
//通过key的hash值获取到的下标，找到的桶里面是空的
//这里隐含的一个步骤是：p被赋值为key对应hash下标存放的node数据，i被赋值为桶下标
tab[i] = newNode(hash, key, value, null);
else {
//table 不为空，桶不为空
Node<K,V> e; K k;
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
//key重复
e = p;
else if (p instanceof TreeNode)
//桶里面的数据结构是红黑树
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
//桶里面的数据结构是链表
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
//到达链表末尾
p.next = newNode(hash, key, value, null);
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
//链表长度超过8，转化为红黑树
treeifyBin(tab, hash);
break;
}
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
//链表后续的数据中，有key值重复的
break;
//p指向p.next
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;
}
以上的关键步骤，我都加上了个人的理解，总体步骤如下：
1. 容器的table为空时，调用resize()，给table分配空间
2. 通过key的hash值，计算桶下标值i，
3. 通过i获取容器的桶，
如果为空，直接new node；
如果不为空，比较桶顶部元素的key值
如果key值重复，直接用新的value覆盖旧value，
如果key不重复，并且节点类型是树节点，则转到树的处理函数
如果key不重复，并且节点类型是不是树节点，遍历节点链表，
如果有重复的key值，则用新value替代旧value，
如果没有重复的key值，在桶尾部添加new node，然后根据桶的大小，如果超出8，则把链表重构成红黑树
看完put函数，接下来get的函数就比较好理解了：
/**
* 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 get(Object key) {
Node<K,V> e;
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) {
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
if ((e = first.next) != null) {
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
}
}
return null;
}
类似put，也是先计算key的hash值，然后调用其他函数来执行真正的get操作。
get的大概步骤如下：
当table不为空，并且通过key的hash计算出来的i值，获取到的桶也不为空时，执行以下步骤，否则返回null
判断桶的首元素，是否跟key相等，相等则返回该数据的value值
如果key不相等，
如果桶的数据结构是红黑树，遍历数结构获取数据
如果桶的数据结构是链表，遍历链表获取数据
注意，无论是get还是put，其key是否相等的判断如下：
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
即（aka，also known as），key的hash值相等，并且key值本身也要相等。而其中，hash值的计算又依赖于其hashCode函数，因此，如果要自定义容器的key的对象，则必须要实现该对象的hashCode函数和equals函数，从而定义其相等的判断条件。