HashMap源码解读
文章目录
基于JDK1.8的hashMap源码
HashMap内部结构
先思考HashMap的作用：存储键值对，存储数据的
再思考计算机中存储数据的方式/结构（数据结构）
常见的数据结构：数组，链表，树形等…
猜测HashMap的数据结构是 数据 + 链表 的形式
那么在代码中：
数组的表示：XXXX[] ——> Node table[]
单向链表的表示：
Class Node{
key
value
Node next
}
于是查看源码
1.可以看到一个Node类型的数组table
/**
* 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;
2.看到一个Node<K,V>对象–（可以表示单向链表）：
/**
* Basic hash bin node, used for most entries. (See below for
* TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
*/
static class Node<K,V> implements Map.Entry<K,V> {
final int hash;
final K key;
V value;
Node<K,V> next;
Node(int hash, K key, V value, Node<K,V> next) {
this.hash = hash;
this.key = key;
this.value = value;
this.next = next;
}
public final K getKey() { return key; }
public final V getValue() { return value; }
public final String toString() { return key + "=" + value; }
public final int hashCode() {
return Objects.hashCode(key) ^ Objects.hashCode(value);
}
public final V setValue(V newValue) {
V oldValue = value;
value = newValue;
return oldValue;
}
public final boolean equals(Object o) {
if (o == this)
return true;
if (o instanceof Map.Entry) {
Map.Entry<?,?> e = (Map.Entry<?,?>)o;
if (Objects.equals(key, e.getKey()) &&
Objects.equals(value, e.getValue()))
return true;
}
return false;
}
}
3.再查看其他属性，先进行一个了解
/**
* The default initial capacity - MUST be a power of two.
*/
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16
数组的默认大小 16
/**
* The load factor used when none specified in constructor.
*/
static final float DEFAULT_LOAD_FACTOR = 0.75f;
数组大小可能不够用 – 扩容
什么时候进行扩容？
16 * 0.75=12 数组扩容的一个标准，当数组容量达到长度*阈值的值后进行扩容
/**
* The bin count threshold for using a tree rather than list for a
* bin. Bins are converted to trees when adding an element to a
* bin with at least this many nodes. The value must be greater
* than 2 and should be at least 8 to mesh with assumptions in
* tree removal about conversion back to plain bins upon
* shrinkage.
*/
static final int TREEIFY_THRESHOLD = 8;
链表的长度也不能无限大
当链表长度超过8时，转换成红黑树结构
/**
* The bin count threshold for untreeifying a (split) bin during a
* resize operation. Should be less than TREEIFY_THRESHOLD, and at
* most 6 to mesh with shrinkage detection under removal.
*/
static final int UNTREEIFY_THRESHOLD = 6;
当红黑树节点小于6时，将红黑树转成链表
/**
* The number of key-value mappings contained in this map.
*/
transient int size;
记录以下数组使用格子的数量
添加元素
putVal方法的原代码：
/**
* 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)
n = (tab = resize()).length;
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
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)
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);
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;
}
}
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;
}
看源码：
1.数组的初始化 – resize()方法
/**
* Initializes or doubles table size. If null, allocates in
* accord with initial capacity target held in field threshold.
* Otherwise, because we are using power-of-two expansion, the
* elements from each bin must either stay at same index, or move
* with a power of two offset in the new table.
*
* @return the table
*/
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) {
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
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);
}
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"})
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;
}
用两个变量记录数组的大小以及 新的扩容阈值大小
将扩容阈值赋值给threashold
初始化数组
2.添加元素之前情况
(1)数组原本的位置为空
直接添加即可
(2)数组原本的位置不为空，且下面是链表的结构
循环遍历判断下一个节点是否为空，若不为空则添加到其下一个节点
(3)数组原本的位置不为空，且下面是红黑树的结构
则继续向下添加红黑树节点
3.添加元素的下标如何确定
生产出一个算法，需要满足如下要求
(1) Int类型
通过对key值取hashCode()即可实现，Node对象中会存储计算出的hash值
(2) 0-15范围：数组大小的范围内
将计算出的hash值对16(数组大小)取模即可实现
源码中：
将 n-1&hash =等价于=> 取模运算
hash值 – 32位的二进制表示
例如数组大小为16
则二进制表示为 000000000000000000000000010000
减一的二进制为 000000000000000000000000001111
与key的hash值取余，则前28位肯定为0，后四位最大为1111–15，最小为0000–0，即取值范围为0-15
所以数组长度必须为2的幂次方，才能在减去一的时候，确定最大值
(3) 尽可能充分利用数组的每一个位置
根据上述所知，数组下标通过 n-1&hash 确定，
n-1已经确定，所以为了结果尽可能分散，就是这个hash值尽可能不同
计算hash值的方法：
/**
* 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);
}
将hashCode进行了位移运算
将高16位和低16位进行了异或运算，这样结果就尽可能不同了
数组的扩容
1.数组的扩容 – resize()方法
必须是2的倍数扩容
将数组的容量进行位移操作，同时将阈值也进行位移操作
即16 ——>32 ， 12 ——>24
2.扩容后，原链表节点如何进行挪动
(1)非红黑树结构
循环遍历将链表重新打散
将原来key的hash与旧的数组长度(并非n-1)进行取余
如果为 0 ，保持在原来的位置不同
如果不为0，加上原来的capacity
原数组长度 000000000000000000000000010000
只有key的hash值 000100100011000000100000000110 的这个位置为0时，结果才会为0
所以这个值只可能有两个地方，一个是原下表的位置，另一个是在下标为（原下表+原容量）的位置
(2)红黑树结构
将原来的红黑树进行split切割，重新定位
总结
1.hashMap底层使用哈希表（数组+链表），当链表过长时会将链表转成红黑树以实现O(logn)时间复杂度内查找
2.hashMap的put过程
(1).对Key求hash值，然后计算下标
(2).如果没有碰撞，直接放入桶中
(3).如果碰撞了，以链表的方式直接添加到后面
(4).如果链表长度超过阈值（TREEIFY_THRESHOLD = 8），就将链表转成红黑树
(5).如果节点已经存在就替换旧值
(6).如果桶满了（容量*加载因子），就需要扩容resize
3.hash函数
(1)将高16位和低16位做一个异或运算
(2) (n-1)&hashCode得到下标
4.扩容机制
(1)容量扩充为原阿里的两倍，然后对每个节点重新计算哈希值
(2)这个值只可能存在两个地方，一个是原下标位置，另一个是在下标为(原下表+原容量)的位置
(3)当红黑树节点小于6(UNTREEIFY_THRESHOLD = 6)时，将红黑树转成链表形式
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