Introduction
HashMap first appeared in JDK1.2, and the underlying implementation is based on the hash algorithm. HashMap allows null keys and null values. It is a non-thread-safe class and may cause problems in a multi-threaded environment.
HashMap data structure in version 1.8:
Why are some linked lists and some red-black trees?
When the default linked list length is greater than 8, it is converted to a tree.
structure
Node is a static inner class in HhaspMap:
//Node是单向链表，实现了Map.Entry接口
static class Node<K,V> implements Map.Entry<K,V> {
final int hash;
final K key;
V value;
Node<K,V> next;
//构造函数
Node(int hash, K key, V value, Node<K,V> next) {
this.hash = hash;
this.key = key;
this.value = value;
this.next = next;
}
// getter and setter ... toString ...
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;
}
}
TreeNode is a red-black tree data structure.
static final class TreeNode<K,V> extends LinkedHashMap.Entry<K,V> {
TreeNode<K,V> parent; // red-black tree links
TreeNode<K,V> left;
TreeNode<K,V> right;
TreeNode<K,V> prev; // needed to unlink next upon deletion
boolean red;
TreeNode(int hash, K key, V val, Node<K,V> next) {
super(hash, key, val, next);
}
/**
* Returns root of tree containing this node.
*/
final TreeNode<K,V> root() {
for (TreeNode<K,V> r = this, p;;) {
if ((p = r.parent) == null)
return r;
r = p;
}
}
class definition
public class HashMap<K,V> extends AbstractMap<K,V>
implements Map<K,V>, Cloneable, Serializable
variable
/**
* 默认初始容量16(必须是2的幂次方)
*/
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4;
/**
* 最大容量，2的30次方
*/
static final int MAXIMUM_CAPACITY = 1 << 30;
/**
* 默认加载因子，用来计算threshold
*/
static final float DEFAULT_LOAD_FACTOR = 0.75f;
/**
* 链表转成树的阈值，当桶中链表长度大于8时转成树
threshold = capacity * loadFactor
*/
static final int TREEIFY_THRESHOLD = 8;
/**
* 进行resize操作时，若桶中数量少于6则从树转成链表
*/
static final int UNTREEIFY_THRESHOLD = 6;
/**
* 桶中结构转化为红黑树对应的table的最小大小
当需要将解决 hash 冲突的链表转变为红黑树时，
需要判断下此时数组容量，
若是由于数组容量太小（小于 MIN_TREEIFY_CAPACITY ）
导致的 hash 冲突太多，则不进行链表转变为红黑树操作，
转为利用 resize() 函数对 hashMap 扩容
*/
static final int MIN_TREEIFY_CAPACITY = 64;
/**
保存Node<K,V>节点的数组
该表在首次使用时初始化，并根据需要调整大小。 分配时，
长度始终是2的幂。
*/
transient Node<K,V>[] table;
/**
* 存放具体元素的集
*/
transient Set<Map.Entry<K,V>> entrySet;
/**
* 记录 hashMap 当前存储的元素的数量
*/
transient int size;
/**
* 每次更改map结构的计数器
*/
transient int modCount;
/**
* 临界值 当实际大小(容量*填充因子)超过临界值时，会进行扩容
*/
int threshold;
/**
* 负载因子：要调整大小的下一个大小值（容量*加载因子）。
*/
final float loadFactor;
Construction method
/**
* 传入初始容量大小，使用默认负载因子值 来初始化HashMap对象
*/
public HashMap(int initialCapacity) {
this(initialCapacity, DEFAULT_LOAD_FACTOR);
}
/**
* 默认容量和负载因子
*/
public HashMap() {
this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
}
/**
* 传入初始容量大小和负载因子 来初始化HashMap对象
*/
public HashMap(int initialCapacity, float loadFactor) {
// 初始容量不能小于0，否则报错
if (initialCapacity < 0)
throw new IllegalArgumentException("Illegal initial capacity: " +
initialCapacity);
// 初始容量不能大于最大值，否则为最大值
if (initialCapacity > MAXIMUM_CAPACITY)
initialCapacity = MAXIMUM_CAPACITY;
//负载因子不能小于或等于0，不能为非数字
if (loadFactor <= 0 || Float.isNaN(loadFactor))
throw new IllegalArgumentException("Illegal load factor: " +
loadFactor);
// 初始化负载因子
this.loadFactor = loadFactor;
// 初始化threshold大小
this.threshold = tableSizeFor(initialCapacity);
}
/**
* 找到大于或等于 cap 的最小2的整数次幂的数。
*/
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;
}
Detailed explanation of tableSizeFor method:
Use bitwise operations to find the number that is greater than or equal to the smallest integer power of 2 of cap. For example, if 10 is used, 16 will be returned.
-
The purpose of letting cap-1 be assigned to n is to make the found target value greater than or equal to the original value. For example, in binary 0100, the decimal value is 4. If you operate directly without subtracting 1, the answer is0001 000016 in decimal system, which is obviously not in line with expectations. -
Shift n right by 1 bit: 001xx…xxx, or bitwise or: 011xx…xxx -
Shift n right by 2 bits: 00011…xxx, or bitwise or: 01111…xxx -
Shift n right by 4 bits... -
Shift n right by 8 bits... -
Shift n to the right by 16 bits. Because int is the largest, 2^32move 1, 2, 4, 8, and 16 bits and take the bitwise OR, which will change all the bits after the 1 in the highest bit to 1. -
Let the result n+1, that is, the value of an integer power of 2 is obtained.
Comes with an example:
loadFactor load factor
For HashMap, load factor is a very important parameter, which reflects the usage of HashMap bucket array. By adjusting the load factor, the time and space complexity of HashMap can be different.
当我们调低负载因子时，HashMap 所能容纳的键值对数量变少。扩容时，重新将键值对存储新的桶数组里，键的键之间产生的碰撞会下降，链表长度变短。此时，HashMap 的增删改查等操作的效率将会变高，这里是典型的拿空间换时间。
相反，如果增加负载因子（负载因子可以大于1），HashMap 所能容纳的键值对数量变多，空间利用率高，但碰撞率也高。这意味着链表长度变长，效率也随之降低，这种情况是拿时间换空间。至于负载因子怎么调节，这个看使用场景了。
一般情况下，我们用默认值就可以了。大多数情况下0.75在时间跟空间代价上达到了平衡所以不建议修改。
查找
public V get(Object key) {
Node<K,V> e;
return (e = getNode(hash(key), key)) == null ? null : e.value;
}
// 获取hash值
static final int hash(Object key) {
int h;
// 拿到key的hash值后与其五符号右移16位取与
// 通过这种方式，让高位数据与低位数据进行异或，以此加大低位信息的随机性，变相的让高位数据参与到计算中。
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
final Node<K,V> getNode(int hash, Object key) {
Node<K,V>[] tab;
Node<K,V> first, e;
int n; K k;
// 定位键值对所在桶的位置
if ((tab = table) != null && (n = tab.length) > 0 &&
(first = tab[(n - 1) & hash]) != null) {
// 判断桶中第一项(数组元素)相等
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
// 桶中不止一个结点
if ((e = first.next) != null) {
// 是否是红黑树，是的话调用getTreeNode方法
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);
// 不是红黑树的话，在链表中遍历查找
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
}
}
return null;
}
注意：
-
HashMap的hash算法( hash()方法)。 -
(n - 1) & hash等价于对 length 取余。
添加
public V put(K key, V value) {
// 调用hash(key)方法来计算hash
return putVal(hash(key), key, value, false, true);
}
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab;
Node<K,V> p;
int n, i;
// 容量初始化：当table为空，则调用resize()方法来初始化容器
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
//确定元素存放在哪个桶中，桶为空，新生成结点放入桶中
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
// 比较桶中第一个元素(数组中的结点)的hash值相等，key相等
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
//如果键的值以及节点 hash 等于链表中的第一个键值对节点时，则将 e 指向该键值对
e = p;
// 如果桶中的引用类型为 TreeNode，则调用红黑树的插入方法
else if (p instanceof TreeNode)
// 放入树中
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
//对链表进行遍历，并统计链表长度
for (int binCount = 0; ; ++binCount) {
// 到达链表的尾部
if ((e = p.next) == null) {
//在尾部插入新结点
p.next = newNode(hash, key, value, null);
// 如果结点数量达到阈值，转化为红黑树
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
treeifyBin(tab, hash);
break;
}
// 判断链表中结点的key值与插入的元素的key值是否相等
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
//判断要插入的键值对是否存在 HashMap 中
if (e != null) { // existing mapping for key
V oldValue = e.value;
// onlyIfAbsent 表示是否仅在 oldValue 为 null 的情况下更新键值对的值
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
// 键值对数量超过阈值时，则进行扩容
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}
事实上，new HashMap();完成后，如果没有put操作，是不会分配存储空间的。
-
当桶数组 table 为空时，通过扩容的方式初始化 table -
查找要插入的键值对是否已经存在，存在的话根据条件判断是否用新值替换旧值 -
如果不存在，则将键值对链入链表中，并根据链表长度决定是否将链表转为红黑树 -
判断键值对数量是否大于阈值，大于的话则进行扩容操作
扩容机制
在 HashMap 中，桶数组的长度均是2的幂，阈值大小为桶数组长度与负载因子的乘积。当 HashMap 中的键值对数量超过阈值时，进行扩容。
HashMap 按当前桶数组长度的2倍进行扩容，阈值也变为原来的2倍（如果计算过程中，阈值溢出归零，则按阈值公式重新计算）。扩容之后，要重新计算键值对的位置，并把它们移动到合适的位置上去。
final Node<K,V>[] resize() {
// 拿到数组桶
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
int oldThr = threshold;
int newCap, newThr = 0;
// 如果数组桶的容量大与0
if (oldCap > 0) {
// 如果比最大值还大，则赋值为最大值
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
// 如果扩容后小于最大值 而且 旧数组桶大于初始容量16， 阈值左移1(扩大2倍)
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
}
// 如果数组桶容量<=0 且 旧阈值 >0
else if (oldThr > 0) // initial capacity was placed in threshold
// 新容量=旧阈值
newCap = oldThr;
// 如果数组桶容量<=0 且 旧阈值 <=0
else { // zero initial threshold signifies using defaults
// 新容量=默认容量
newCap = DEFAULT_INITIAL_CAPACITY;
// 新阈值= 负载因子*默认容量
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
}
// 如果新阈值为0
if (newThr == 0) {
// 重新计算阈值
float ft = (float)newCap * loadFactor;
newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
(int)ft : Integer.MAX_VALUE);
}
// 更新阈值
threshold = newThr;
@SuppressWarnings({
"rawtypes","unchecked"})
// 创建新数组
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
// 覆盖数组桶
table = newTab;
// 如果旧数组桶不是空，则遍历桶数组，并将键值对映射到新的桶数组中
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;
// 如果是红黑树
else if (e instanceof TreeNode)
// 重新映射时，需要对红黑树进行拆分
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
// 如果不是红黑树，则按链表处理
Node<K,V> loHead = null, loTail = null;
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
// 遍历链表，并将链表节点按原顺序进行分组
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);
// 将分组后的链表映射到新桶中
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}
整体步骤：
-
计算新桶数组的容量 newCap 和新阈值 newThr -
根据计算出的 newCap 创建新的桶数组，桶数组 table 也是在这里进行初始化的 -
将键值对节点重新映射到新的桶数组里。如果节点是 TreeNode 类型，则需要拆分红黑树。如果是普通节点，则节点按原顺序进行分组。
总结起来，一共有三种扩容方式：
-
使用默认构造方法初始化HashMap。从前文可以知道HashMap在一开始初始化的时候会返回一个空的table，并且thershold为0。因此第一次扩容的容量为默认值 DEFAULT_INITIAL_CAPACITY也就是16。同时threshold = DEFAULT_INITIAL_CAPACITY * DEFAULT_LOAD_FACTOR = 12。 -
指定初始容量的构造方法初始化 HashMap。那么从下面源码可以看到初始容量会等于threshold，接着threshold = 当前的容量（threshold） * DEFAULT_LOAD_FACTOR。 -
HashMap不是第一次扩容。如果 HashMap已经扩容过的话，那么每次table的容量以及threshold量为原有的两倍。
细心点的人会很好奇，为什么要判断loadFactor为0呢？
loadFactor小数位为 0，整数位可被2整除且大于等于8时，在某次计算中就可能会导致 newThr 溢出归零。
疑问和进阶
1. JDK1.7是基于数组+单链表实现（为什么不用双链表）
首先，用链表是为了解决hash冲突。
单链表能实现为什么要用双链表呢?(双链表需要更大的存储空间)
2. 为什么要用红黑树，而不用平衡二叉树？
插入效率比平衡二叉树高，查询效率比普通二叉树高。所以选择性能相对折中的红黑树。
3. 重写对象的Equals方法时，要重写hashCode方法，为什么？跟HashMap有什么关系？
equals与hashcode间的关系:
-
If the two objects are the same (that is, comparing with equals and returning true), then their hashCode values must be the same; -
If the hashCode of two objects is the same, they are not necessarily the same (that is, comparing with equals returns false)
Because when traversing and judging in the linked list structure of HashMap, the business logic of the rewritten equals method to compare whether objects are equal under certain circumstances is more complicated, and looping it will affect the search efficiency. So the hashcode judgment is put first here. As long as the hashcodes are not equal, the game is over, and there is no need to call complicated equals. Improves the usage efficiency of HashMap to a great extent.
Therefore, the hashcode method is rewritten so that we can use collection classes such as HashMap normally, because HashMap requires both hashcode and equals comparison to determine whether objects are equal. This implementation is to improve the efficiency of HashMap.
Attached is the source code picture:
4. Why doesn’t HashMap directly use the original hash value of the object?
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
We found that the hash value of HashMap is obtained through the above method, not through key.hashCode()the method.
reason:
Through shifting and XOR operations, the hash can be made more complex, thereby affecting the distribution of the hash.
5. Since red-black trees are so good, why doesn’t hashmap directly use red-black trees, but only converts to red-black trees when there are more than 8?
Because the red-black tree requires left-turn and right-turn operations, but the singly linked list does not.
The following are comparisons of singly linked lists and red-black tree structures.
If there are less than 8 elements, the query cost is high and the new cost is low.
If there are more than 8 elements, the query cost is low and the new cost is high.
As for why the number 8 was chosen, it was the result of a compromise by the boss -.-, just like the default value of loadFactor of 0.75.
This article is published by mdnice multi-platform