layout: post title: "java之HashMap" subtitle: " "1.底层实现原理 2.源码分析"" date: 2018-11-11 07:00:00 author: "青乡" header-img: "img/post-bg-2015.jpg" catalog: true tags: - DataStructure&Algorithm - HashMap
数组是连续存储吗？
转换流程
key——》hashcode——》数组索引i。
存储是否连续？
[null, null, null, aaa=111, null, null, null, null, null, null, null, null, null, null, bbb=111, null]
不连续。只是计算出来的数组索引值肯定是在数组大小范围之内，1.不连续；2.具体是哪个位置，看计算出来的索引值。
为什么计算出来的数组索引值是在数组大小范围之内？具体是如何计算和如何确保的？
/**
* Returns index for hash code h.
*/
static int indexFor(int h, int length) {
// assert Integer.bitCount(length) == 1 : "length must be a non-zero power of 2";
return h & (length-1); //按位与
}
索引值是根据hashcode和数组长度做按位与计算得到的，按位与的关键是都比较小的那个数据(即数组长度)，因为按位与的结果肯定比比较小的数据还要小(因为同时为1才为1)，所以按位与的结果肯定是在数组长度范围之内。
如何解决冲突？
有2种情况可能出现冲突，
1.不同的key，hashcode一样，所以索引值肯定也是一样。这里冲突的关键是hash算法。
2.不同的key，hashcode不一样，但是按位与得到的索引值一样。有没有这种情况？
HashMap如何解决冲突？
1.数组
2.链表
如果冲突，就把冲突对象覆盖旧的对象，然后新的对象指向旧的对象，形成一个链表，这也是节点Entry的数据结构以及为什么要使用链表这种数据结构。
在HashMap里，数组的索引又叫桶bucket，就是说某个数组元素指向的对象可能是一个对象，也可能是一个链表(出现冲突的时候)。
xiaolu123456.iteye.com/blog/148534…
key一样，会覆盖数据吗？
会。
怎么覆盖？
/**
* 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) {
if (table == EMPTY_TABLE) {
inflateTable(threshold);
}
if (key == null)
return putForNullKey(value);
int hash = hash(key);
int i = indexFor(hash, table.length);
for (Entry<K,V> e = table[i]; e != null; e = e.next) {
Object k;
if (e.hash == hash && ((k = e.key) == key || key.equals(k))) { //如果key一样，新的value覆盖旧的value
V oldValue = e.value;
e.value = value;
e.recordAccess(this);
return oldValue;
}
}
modCount++;
addEntry(hash, key, value, i);
return null;
}
容量为什么是2的倍数？
1.初始容量是2的倍数16
因为容量是移位运算。左移一位就是2~2=4，左移4位就是2~4=16。
/**
* The default initial capacity - MUST be a power of two.
*/
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16 //初始大小是16
2.每次扩容乘以2
/**
* Adds a new entry with the specified key, value and hash code to
* the specified bucket. It is the responsibility of this
* method to resize the table if appropriate.
*
* Subclass overrides this to alter the behavior of put method.
*/
void addEntry(int hash, K key, V value, int bucketIndex) {
if ((size >= threshold) && (null != table[bucketIndex])) {
resize(2 * table.length); //容量不够，扩容乘以2
hash = (null != key) ? hash(key) : 0;
bucketIndex = indexFor(hash, table.length);
}
createEntry(hash, key, value, bucketIndex);
}
源码分析
put()方法
//jdk7-put()方法
/**
* 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) {
if (table == EMPTY_TABLE) {
inflateTable(threshold);
}
if (key == null)
return putForNullKey(value);
int hash = hash(key);
int i = indexFor(hash, table.length);
for (Entry<K,V> e = table[i]; e != null; e = e.next) {
Object k;
if (e.hash == hash && ((k = e.key) == key || key.equals(k))) {
V oldValue = e.value;
e.value = value;
e.recordAccess(this);
return oldValue;
}
}
modCount++;
addEntry(hash, key, value, i);
return null;
}
/**
* An empty table instance to share when the table is not inflated.
*/
static final Entry<?,?>[] EMPTY_TABLE = {}; //数组初始值是空
/**
* The table, resized as necessary. Length MUST Always be a power of two.
*/
transient Entry<K,V>[] table = (Entry<K,V>[]) EMPTY_TABLE;
/**
* The next size value at which to resize (capacity * load factor).
* @serial
*/
// If table == EMPTY_TABLE then this is the initial capacity at which the
// table will be created when inflated.
int threshold;
jdk版本？ //jdk版本？
/**
* 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) //hashcode, key //key, value //value, false, true);
}
/**
* 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;
}
//hash()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); //调用String.hashCode()
}
//底层是数组
/**
* 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; //节点数组
//初始化数组大小或double扩容数组大小
/**
* 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.
*
* @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;
}
get()方法
/**
* 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) {
if (key == null)
return getForNullKey();
Entry<K,V> entry = getEntry(key);
return null == entry ? null : entry.getValue();
}
/**
* Returns the entry associated with the specified key in the
* HashMap. Returns null if the HashMap contains no mapping
* for the key.
*/
final Entry<K,V> getEntry(Object key) {
if (size == 0) {
return null;
}
int hash = (key == null) ? 0 : hash(key);
for (Entry<K,V> e = table[indexFor(hash, table.length)];
e != null;
e = e.next) { //读数据的有2种情况：1.无冲突，for一次就结束，因为已经读到数据 2.如果有冲突，就要遍历链表，直到找到key和hashcode都相同的那个节点
Object k;
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
}
return null;
}
直接地址法(线性函数)和链地址法(使用链表)？
hash算法，有2种：
1.线性函数
2.使用链表
作用
不管是哪种hash算法，目的是避免冲突问题，如果出现冲突，则要解决冲突问题。
线性函数
f(key) = a × key + b
就是key和hashcode存在一个简单的固定的线性函数关系。而且不会发生冲突。但是较少使用，因为需要事先知道key和hashcode的这么一种对应的线性函数关系，而实际应用中都不知道也没有这种线性函数关系。
使用链表
节点使用链表这种数据结构。目的是为了解决冲突问题。
HashMap的节点就是使用链表来解决冲突问题的。
