Java HashMap 源码分析
HashMap实现简介
HashMap 底层采用节点数组,数组内存储的是链表或者红黑树(JDK8)
1. 源码分析
1.1 属性
/**
* The default initial capacity - MUST be a power of two.
* 默认容量必须是2的倍数 这里是16
*/
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16
/**
* The maximum capacity, used if a higher value is implicitly specified
* by either of the constructors with arguments.
* MUST be a power of two <= 1<<30.
* 这里是2的30次方,int正数的大小最大是2的31次方
*/
static final int MAXIMUM_CAPACITY = 1 << 30;
/**
* The load factor used when none specified in constructor.
* 装载系数, 最大容量=容量*装载系数
*/
static final float DEFAULT_LOAD_FACTOR = 0.75f;
/**
* 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.
*
* 链表的最大长度是8，当超过这个长度时转换成树
*
*/
static final int TREEIFY_THRESHOLD = 8;
/**
* 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.
*
*当执行resize操作时，当容器中bin的数量少于UNTREEIFY_THRESHOLD时使用链表来代替树。默认值是6
*/
static final int UNTREEIFY_THRESHOLD = 6;
/**
* The smallest table capacity for which bins may be treeified.
* (Otherwise the table is resized if too many nodes in a bin.)
* Should be at least 4 * TREEIFY_THRESHOLD to avoid conflicts
* between resizing and treeification thresholds.
*
* 当桶中的bin被树化时最小的hash表容量。（如果没有达到这个阈值，即hash表容量小于
* MIN_TREEIFY_* CAPACITY，当桶中bin的数量太多时会执行resize扩容操作）
* 这个MIN_TREEIFY_CAPACITY的值至少是TREEIFY_THRESHOLD的4倍
*/
static final int MIN_TREEIFY_CAPACITY = 64;
/**
* 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.)
* 节点数组
*/
transient Node<K,V>[] table;1.2 链表节点实现
static class Node<K,V> implements Map.Entry<K,V> {
final int hash; //hash值
final K key; //key值
V value; //value
Node<K,V> next; //下一个索引
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; }
/**
* 获取节点的hash值,key、value的hash值求异或
*
**/
public final int hashCode() {
return Objects.hashCode(key) ^ Objects.hashCode(value);
}
public final V setValue(V newValue) {
V oldValue = value;
value = newValue;
return oldValue;
}
//节点key,vaule值都相等，节点才相等
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;
}
}1.3 构造函数
/**
* Constructs an empty <tt>HashMap</tt> with the specified initial
* capacity and load factor.
*
* @param initialCapacity the initial capacity
* @param loadFactor the load factor
* @throws IllegalArgumentException if the initial capacity is negative
* or the load factor is nonpositive
*/
public HashMap(int initialCapacity, float loadFactor) {
if (initialCapacity < 0)
throw new IllegalArgumentException("Illegal initial capacity: " +
initialCapacity);
//初始值大小
if (initialCapacity > MAXIMUM_CAPACITY)
initialCapacity = MAXIMUM_CAPACITY;
if (loadFactor <= 0 || Float.isNaN(loadFactor))
throw new IllegalArgumentException("Illegal load factor: " +
loadFactor);
//装载因子
this.loadFactor = loadFactor;
//阙值
this.threshold = tableSizeFor(initialCapacity);
}
/**
* Returns a power of two size for the given target capacity.
*/
static final int tableSizeFor(int cap) {
int n = cap - 1;
n |= n >>> 1;
n |= n >>> 2;
n |= n >>> 4;
n |= n >>> 8;
n |= n >>> 16;
return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
}tableSizeFor的功能：
调用 int result = tableSizeFor(18);
- int n = cap -1;
n = 17 二进制 n = 10001
- n |= n >>> 1;
10001 n
| 01000 n >>> 1
---------------
11001- n |= n >>> 2;
11001 n
| 00110 n >>> 2
---------------
11111- n |= n >>> 4;
11111 n
| 00001 n >>> 4
---------------
11111- n |= n >>> 4;
11111 n
| 00000 n >>> 16
---------------
11111 n = 31java int占有4个字节32位,上述计算n总共无符号右移32位 n+1 = 32
由上分析tableSizeFor(int cap)方法返回大于等于cap的最小2的幂值
tableSizeFor(31) –> 32
tableSizeFor(33) –> 64
1.4 put方法
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
/**
* 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.
*
* key值的hash值与hash值右移16位求异或，作为key的hash值
*/
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
/**
* 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)
//tab为空,调整table的大小
n = (tab = resize()).length;
//tab索引的计算方式 tab的大小与hash求与
if ((p = tab[i = (n - 1) & hash]) == null)
//当前index位置不存在元素,新分配节点
tab[i] = newNode(hash, key, value, null);
else {
//index位置已存在节点
Node<K,V> e; K k;
//index位置已存在首节点的hash值与put进来的hash值一样
//并且节点key值与新节点key值相同
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
//index已存在首节点,且是TreeNode类型,树容器put
else if (p instanceof TreeNode)
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
//其他,普通情况
else {
for (int binCount = 0; ; ++binCount) {
//新put的节点链接到数组索引处链表的最后一个位置
if ((e = p.next) == null) {
p.next = newNode(hash, key, value, null);
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
//链表长度超出转换为树结构的阙值,树型转换
treeifyBin(tab, hash);
break;
}
//index处的链表某节点与put的新节点相同
//节点hash相同,key相同或值相等
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
//value赋值
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
//table元素大小超过阙值,重新分配大小
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}
1.4.1 hash表的存储方式
Node
graph LR
A-->A1
A1-->A2
A2-->A3
B-->B1
B1-->B2
B2-->B3
C-->C1
C1-->C2
C2-->C3
1.4.2 key 的索引计算方式
index = (tables.length - 1) & [(hash = key.hashCode()) ^ (hash >>> 16)]
tables的大小只能是2的幂 16,32,64 …..
tables.length -1 的二进制数也就是全1, 1111,11111,111111
其与任何二进制数求与得到index值,则index都会小于tables.length
1.4.3 散列表冲突解决
链接方法
hashMap采用链表的方式解决hash冲突
当hash函数求得的index与已经存在的rootNode相同时,如果key也相同则覆盖其value值,key值不相同的话进行此index处的链表遍历,其中发现key值存在的话进行覆盖,不存在的话新node附加到链表尾部- 开放寻址法
开放寻址发，所有的元素都被散列到表中,也就是说,每个表项或包含动态集合中的一个元素,或包含NIL,当查找某个元素时候，要系统的检查所有的表项，直到查到所需元素，或元素不在表中
1.5 resize重新分配大小
/**
* 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.
* 重新分配table大小,原大小的二倍
* @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;
}
//数组大小没有大于最大值
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
//阙值也放大二倍
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;
//阙值=容量*装载因子
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;
//注***见下
//不需要调整位置
if ((e.hash & oldCap) == 0) {
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
}
//需要调整位置
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;
}注*
index的计算方式hash & (cap -1)此处为什么是 (e.hash & oldCap) == 0
假设oldCap=16,oldindex = e.hash & 15
e.hash
& 00001111 15
-------------------
0000xxxx 后四位决定索引的值
现在进行扩容 newCap = 32,newIndex = e.hash & 31
e.hash
& 000011111 31
-------------------
0000yxxxx 后五位决定新索引的值,且第五位决定index是否变动为oldCap+index看下e.hash & oldCap的结果:
e.hash
& 000010000 16
-------------------
0000y0000 如果e.hash的后数第五位是0那么结果就是y=0那么节点新索引就没有变化,也就有了上面后面的代码,将索引需要变化的节点组成新的链表,链接到newTable的oldCap+j位置,索引不需要变化的重新组成链表放到原索引处
1.6 红黑树部分
当链表的长度大于阙值时，链表会被转换为红黑树,
链表的搜索、插入等操作时间复杂度O(n),红黑树为平衡二叉树其为lg(n)(最坏情况下也是)