使用数组加链表实现hashmap。
hashmap初始容量为16，装载因子为0.75，hashmap里面的元素达到阈值(阈值=容量 * 装载因子)时，就会进行扩容，每次扩容2倍。
public V put(K key, V value) {
// 当第一次往里面放元素时才真正地申请空间
if (table == EMPTY_TABLE) {
inflateTable(threshold);
}
if (key == null)
return putForNullKey(value);
// 计算hash值
int hash = hash(key);
// 计算索引值，大小为0到table.length - 1
int i = indexFor(hash, table.length);
// 如果已经存在这个key，则更新value值
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++;
// 不存在这个key,就新增
addEntry(hash, key, value, i);
return null;
}为数组申请空间
private void inflateTable(int toSize) {
// Find a power of 2 >= toSize
int capacity = roundUpToPowerOf2(toSize);
threshold = (int) Math.min(capacity * loadFactor, MAXIMUM_CAPACITY + 1);
table = new Entry[capacity];
// initHashSeedAsNeeded方法判断是否需要rehash
initHashSeedAsNeeded(capacity);
}每次最接近number的数，而且是2的幂次方的数。如number=8就返回8，number=9就返回15
private static int roundUpToPowerOf2(int number) {
// assert number >= 0 : "number must be non-negative";
return number >= MAXIMUM_CAPACITY
? MAXIMUM_CAPACITY
: (number > 1) ? Integer.highestOneBit((number - 1) << 1) : 1;
}如果已经存在null键则更新其value，否则将null键添加到数组中
private V putForNullKey(V value) {
for (Entry<K,V> e = table[0]; e != null; e = e.next) {
if (e.key == null) {
V oldValue = e.value;
e.value = value;
e.recordAccess(this);
return oldValue;
}
}
modCount++;
addEntry(0, null, value, 0);
return null;
}将hashcode的高位和低位混合求hash值，减少冲突
final int hash(Object k) {
int h = hashSeed;
if (0 != h && k instanceof String) {
return sun.misc.Hashing.stringHash32((String) k);
}
h ^= k.hashCode();
// This function ensures that hashCodes that differ only by
// constant multiples at each bit position have a bounded
// number of collisions (approximately 8 at default load factor).
h ^= (h >>> 20) ^ (h >>> 12);
return h ^ (h >>> 7) ^ (h >>> 4);
}hash & (length -1)可以将所有hash值映射到0到length-1范围内，这也解释了为什么hashmap的容量必须是2的倍数。
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);
}void addEntry(int hash, K key, V value, int bucketIndex) {
// 扩容
if ((size >= threshold) && (null != table[bucketIndex])) {
resize(2 * table.length);
// 重新hash值
hash = (null != key) ? hash(key) : 0;
// 重新计算索引位置
bucketIndex = indexFor(hash, table.length);
}
// 添加新元素
createEntry(hash, key, value, bucketIndex);
}void resize(int newCapacity) {
// 旧数组
Entry[] oldTable = table;
// 旧容量
int oldCapacity = oldTable.length;
// 如果旧容量等于最大容量，则直接返回，无法扩容
if (oldCapacity == MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return;
}
// 新建一个数组
Entry[] newTable = new Entry[newCapacity];
// 将原始table中元素复制到newTable
transfer(newTable, initHashSeedAsNeeded(newCapacity));
table = newTable;
threshold = (int)Math.min(newCapacity * loadFactor, MAXIMUM_CAPACITY + 1);
}initHashSeedAsNeeded方法判断是否需要rehash
final boolean initHashSeedAsNeeded(int capacity) {
// hashSeed降低hash碰撞的hash种子，初始值为0
boolean currentAltHashing = hashSeed != 0;
//ALTERNATIVE_HASHING_THRESHOLD： 当map的capacity容量大于这个值的时候并满足其他条件时候进行重新hash
boolean useAltHashing = sun.misc.VM.isBooted() && (capacity >= Holder.ALTERNATIVE_HASHING_THRESHOLD);
//TODO 异或操作，二者满足一个条件即可rehash
boolean switching = currentAltHashing ^ useAltHashing;
if (switching) {
// 更新hashseed的值
hashSeed = useAltHashing ? sun.misc.Hashing.randomHashSeed(this) : 0;
}
return switching;
}将原始table中元素复制到newTable
如果多线程并发扩容时会形成循环链表，线程A执行完Entry<K,V> next = e.next;，如果让给线程B执行，线程B执行完扩容后，线程A还会继续扩容，这种情况下就会形成循环链表。采用尾插法可以解决这个问题，所以jdk1.8就采用尾插法了。
void transfer(Entry[] newTable, boolean rehash) {
int newCapacity = newTable.length;
for (Entry<K,V> e : table) {
while(null != e) {
Entry<K,V> next = e.next;
// 如果需要重新hash就重新hash
if (rehash) {
e.hash = null == e.key ? 0 : hash(e.key);
}
// 计算索引位置
int i = indexFor(e.hash, newCapacity);
// 头插法
e.next = newTable[i];
newTable[i] = e;
e = next;
}
}
}如果没有超过阈值，则直接存入table
void createEntry(int hash, K key, V value, int bucketIndex) {
Entry<K,V> e = table[bucketIndex];
// 头插法
table[bucketIndex] = new Entry<>(hash, key, value, e);
size++;
}Entry(int h, K k, V v, Entry<K,V> n) {
value = v;
next = n;
key = k;
hash = h;
}public V get(Object key) {
if (key == null)
return getForNullKey();
// 调用getEntry方法
Entry<K,V> entry = getEntry(key);
return null == entry ? null : entry.getValue();
}key = null映射为索引0，查找table[0]中有没有key=null的结点
private V getForNullKey() {
if (size == 0) {
return null;
}
for (Entry<K,V> e = table[0]; e != null; e = e.next) {
if (e.key == null)
return e.value;
}
return null;
}final Entry<K,V> getEntry(Object key) {
if (size == 0) {
return null;
}
int hash = (key == null) ? 0 : hash(key);
// 先找到在那个table[i]，然后遍历链表找到这个key
for (Entry<K,V> e = table[indexFor(hash, table.length)];
e != null;
e = e.next) {
Object k;
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
}
return null;
}HashMap结构图：
HashMap：它根据键的hashCode值存储数据，大多数情况下可以直接定位到它的值，因而具有很快的访问速度，但遍历顺序却是不确定的。 HashMap最多只允许一条记录的键为null，允许多条记录的值为null。HashMap非线程安全，即任一时刻可以有多个线程同时写HashMap，可能会导致数据的不一致。如果需要满足线程安全，可以用 Collections的synchronizedMap方法使HashMap具有线程安全的能力，或者使用ConcurrentHashMap。
默认的初始容量是16 默认的负载因子0.75 当桶上的结点数大于等于8会转成红黑树 当桶上的结点数小于6红黑树转链表 Node<k,v>[] table //存储元素的哈希桶数组，总是2的幂次倍
public class HashMap<K,V> extends AbstractMap<K,V>
implements Map<K,V>, Cloneable, Serializable {
private static final long serialVersionUID = 362498820763181265L;
/**
* The default initial capacity - MUST be a power of two.
* 默认容量
*/
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.
* 最大容量
*/
static final int MAXIMUM_CAPACITY = 1 << 30;
/**
* The load factor used when none specified in constructor.
* 装载因子
*/
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.
* 当桶中结点达到8时，将转为红黑树
*/
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.
* 当桶中结点小于6时，将由红黑树转为链表
*/
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.
* 桶树化，哈希表最小的容量
*/
static final int MIN_TREEIFY_CAPACITY = 64;
/**
* Basic hash bin node, used for most entries. (See below for
* TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
* 数组中每一个元素的类型
*/
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;
}
}
/* ---------------- Static utilities -------------- */
/**
* 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.
* hash值计算方法，将key.hashCode() 与 key.hashCode()右移16位做异或运算
*/
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
/**
* Returns x's Class if it is of the form "class C implements
* Comparable<C>", else null.
*/
static Class<?> comparableClassFor(Object x) {
if (x instanceof Comparable) {
Class<?> c; Type[] ts, as; Type t; ParameterizedType p;
if ((c = x.getClass()) == String.class) // bypass checks
return c;
if ((ts = c.getGenericInterfaces()) != null) {
for (int i = 0; i < ts.length; ++i) {
if (((t = ts[i]) instanceof ParameterizedType) &&
((p = (ParameterizedType)t).getRawType() ==
Comparable.class) &&
(as = p.getActualTypeArguments()) != null &&
as.length == 1 && as[0] == c) // type arg is c
return c;
}
}
}
return null;
}
/**
* Returns k.compareTo(x) if x matches kc (k's screened comparable
* class), else 0.
*/
@SuppressWarnings({"rawtypes","unchecked"}) // for cast to Comparable
static int compareComparables(Class<?> kc, Object k, Object x) {
return (x == null || x.getClass() != kc ? 0 :
((Comparable)k).compareTo(x));
}
/**
* Returns a power of two size for the given target capacity.
* 返回大于输入参数且最近的2的整数次幂的数。如cap=10,返回结果就是16
*/
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;
}
/* ---------------- Fields -------------- */
/**
* 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.)
* 数组
*/
transient Node<K,V>[] table;
/**
* Holds cached entrySet(). Note that AbstractMap fields are used
* for keySet() and values().
*/
transient Set<Map.Entry<K,V>> entrySet;
/**
* The number of key-value mappings contained in this map.
* map包含的元素大小
*/
transient int size;
/**
* The number of times this HashMap has been structurally modified
* Structural modifications are those that change the number of mappings in
* the HashMap or otherwise modify its internal structure (e.g.,
* rehash). This field is used to make iterators on Collection-views of
* the HashMap fail-fast. (See ConcurrentModificationException).
* 修改次数
*/
transient int modCount;
/**
* The next size value at which to resize (capacity * load factor).
*
* @serial
*/
// (The javadoc description is true upon serialization.
// Additionally, if the table array has not been allocated, this
// field holds the initial array capacity, or zero signifying
// DEFAULT_INITIAL_CAPACITY.)
// 阈值就是容量*装载因子
int threshold;
/**
* The load factor for the hash table.
*
* @serial
*/
final float loadFactor;
/* ---------------- Public operations -------------- */
/**
* 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);
if (initialCapacity > MAXIMUM_CAPACITY)
initialCapacity = MAXIMUM_CAPACITY;
if (loadFactor <= 0 || Float.isNaN(loadFactor))
throw new IllegalArgumentException("Illegal load factor: " +
loadFactor);
this.loadFactor = loadFactor;
// 找到最接近initialCapacity的，且是2的幂次方的数
this.threshold = tableSizeFor(initialCapacity);
}
/**
* Constructs an empty <tt>HashMap</tt> with the specified initial
* capacity and the default load factor (0.75).
*
* @param initialCapacity the initial capacity.
* @throws IllegalArgumentException if the initial capacity is negative.
*/
public HashMap(int initialCapacity) {
this(initialCapacity, DEFAULT_LOAD_FACTOR);
}
/**
* Constructs an empty <tt>HashMap</tt> with the default initial capacity
* (16) and the default load factor (0.75).
*/
public HashMap() {
this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
}
/**
* Constructs a new <tt>HashMap</tt> with the same mappings as the
* specified <tt>Map</tt>. The <tt>HashMap</tt> is created with
* default load factor (0.75) and an initial capacity sufficient to
* hold the mappings in the specified <tt>Map</tt>.
*
* @param m the map whose mappings are to be placed in this map
* @throws NullPointerException if the specified map is null
*/
public HashMap(Map<? extends K, ? extends V> m) {
this.loadFactor = DEFAULT_LOAD_FACTOR;
putMapEntries(m, false);
}
/**
* Implements Map.putAll and Map constructor
*
* @param m the map
* @param evict false when initially constructing this map, else
* true (relayed to method afterNodeInsertion).
* putMapEntries调用了putVal, putVal已详细注释
*/
final void putMapEntries(Map<? extends K, ? extends V> m, boolean evict) {
int s = m.size();
if (s > 0) {
if (table == null) { // pre-size
float ft = ((float)s / loadFactor) + 1.0F;
int t = ((ft < (float)MAXIMUM_CAPACITY) ?
(int)ft : MAXIMUM_CAPACITY);
if (t > threshold)
threshold = tableSizeFor(t);
}
else if (s > threshold)
// 扩容
resize();
for (Map.Entry<? extends K, ? extends V> e : m.entrySet()) {
K key = e.getKey();
V value = e.getValue();
putVal(hash(key), key, value, false, evict);
}
}
}
/**
* Returns the number of key-value mappings in this map.
*
* @return the number of key-value mappings in this map
*/
public int size() {
return size;
}
/**
* Returns <tt>true</tt> if this map contains no key-value mappings.
*
* @return <tt>true</tt> if this map contains no key-value mappings
*/
public boolean isEmpty() {
return size == 0;
}
/**
* 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;
// table != null, 数组不为空，tab[(n - 1) & hash])存在才查找，否则返回null
if ((tab = table) != null && (n = tab.length) > 0 &&
(first = tab[(n - 1) & hash]) != null) {
// 检查第一个结点是不是就是需要查找的key
if (first.hash == hash && // always check first node
((k = first.key) == key || (key != null && key.equals(k))))
return first;
// 如果第一个结点不是，就继续往下查找，一种是红黑树的情况，一种是链表的情况
if ((e = first.next) != null) {
if (first instanceof TreeNode)
return ((TreeNode<K,V>)first).getTreeNode(hash, key);
// 遍历链表
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
return e;
} while ((e = e.next) != null);
}
}
return null;
}
/**
* Returns <tt>true</tt> if this map contains a mapping for the
* specified key.
*
* @param key The key whose presence in this map is to be tested
* @return <tt>true</tt> if this map contains a mapping for the specified
* key.
* 获取不到这个key，就是不存在
*/
public boolean containsKey(Object key) {
return getNode(hash(key), key) != null;
}
/**
* 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);
}
/**
* 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
* 核心方法之一，存入key、value
*/
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
// 如果table=null或者tab.length=0，就进行扩容
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
// 此处对应没有产生hash冲突的情况，直接使tab[i]等于一个新创建的Node结点
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else { // 产生hash冲突
Node<K,V> e; K k;
// 第一个结点就是要找的结点
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
// 到红黑树中查找
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
// p.next = null表示没有找到，直接插入一个新结点
p.next = newNode(hash, key, value, null);
// 如果binCount大于等于7，就转化为红黑树
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;
}
}
// 更新结点值
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
// 如果大于阈值，就进行扩容
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}
/**
* 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
* 扩容，返回一个新数组
*/
final Node<K,V>[] resize() {
// 旧数组
Node<K,V>[] oldTab = table;
// 旧容量
int oldCap = (oldTab == null) ? 0 : oldTab.length;
// 旧阈值
int oldThr = threshold;
// 新容量，新阈值
int newCap, newThr = 0;
if (oldCap > 0) {
// 旧容量大于等于最大容量，则阈值直接等于int最大值，返回原数组，无法进行扩容了
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
// 新容量设置为旧容量的2倍，新阈值也设置为旧阈值的2倍
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
// 新容量设置为旧阈值
newCap = oldThr;
else { // zero initial threshold signifies using defaults
// 旧容量等于0的情况，说明刚进行插入，使用默认容量
newCap = DEFAULT_INITIAL_CAPACITY;
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
}
// 设置新容量的值
if (newThr == 0) {
float ft = (float)newCap * loadFactor;
newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
(int)ft : Integer.MAX_VALUE);
}
threshold = newThr;
// 扩容就是新建一个数组，把原数组里面的数据存入新数组
@SuppressWarnings({"rawtypes","unchecked"})
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
if (oldTab != null) {
// 下面这个循环，就是将原数组中的数据存到新数组
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
if ((e = oldTab[j]) != null) {
oldTab[j] = null;
// 数组j的位置只有一个元素
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
else if (e instanceof TreeNode)
// 对应红黑树的情况
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
// 链表
// loHead和loTail用于记录oldTab[j]中结点，扩容后索引位置不变的情况
Node<K,V> loHead = null, loTail = null;
// hiHead和hiTail用于记录oldTab[j]中结点，扩容后索引位置等于原位置+原容量的情况
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
do {
next = e.next;
// 这个if成立，说明e结点扩容后还是在j位置
if ((e.hash & oldCap) == 0) {
// 尾插法
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
}
else {
// 这个if成立，说明e结点扩容后，存储在j+oldCap位置，也是尾插法
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
hiTail = e;
}
} while ((e = next) != null);
// loHead链表扩容后还在j位置
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
// hiHead链表对应新数组的位置就是，j + oldCap，这里也就是为什么数组大小一定要是2的倍数
// 注意到，这里不需要重新hash，可以节省hash时间，其实就算就行重新hash，rehash的值也
// 是j + oldCap，这就是hashmap非常巧妙的地方。
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}
/**
* Replaces all linked nodes in bin at index for given hash unless
* table is too small, in which case resizes instead.
* 红黑树太复杂，具体操作就不看了，把其他内容掌握就可以了
*/
final void treeifyBin(Node<K,V>[] tab, int hash) {
int n, index; Node<K,V> e;
if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
resize();
else if ((e = tab[index = (n - 1) & hash]) != null) {
TreeNode<K,V> hd = null, tl = null;
do {
TreeNode<K,V> p = replacementTreeNode(e, null);
if (tl == null)
hd = p;
else {
p.prev = tl;
tl.next = p;
}
tl = p;
} while ((e = e.next) != null);
if ((tab[index] = hd) != null)
hd.treeify(tab);
}
}
/**
* Copies all of the mappings from the specified map to this map.
* These mappings will replace any mappings that this map had for
* any of the keys currently in the specified map.
*
* @param m mappings to be stored in this map
* @throws NullPointerException if the specified map is null
*/
public void putAll(Map<? extends K, ? extends V> m) {
putMapEntries(m, true);
}
/**
* Removes the mapping for the specified key from this map if present.
*
* @param key key whose mapping is to be removed from the map
* @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 remove(Object key) {
Node<K,V> e;
return (e = removeNode(hash(key), key, null, false, true)) == null ?
null : e.value;
}
/**
* Implements Map.remove and related methods
*
* @param hash hash for key
* @param key the key
* @param value the value to match if matchValue, else ignored
* @param matchValue if true only remove if value is equal
* @param movable if false do not move other nodes while removing
* @return the node, or null if none
* 删除结点
*/
final Node<K,V> removeNode(int hash, Object key, Object value,
boolean matchValue, boolean movable) {
Node<K,V>[] tab; Node<K,V> p; int n, index;
if ((tab = table) != null && (n = tab.length) > 0 &&
(p = tab[index = (n - 1) & hash]) != null) {
Node<K,V> node = null, e; K k; V v;
// 第一个结点就是要删除的结点，用node记录下来
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
node = p;
else if ((e = p.next) != null) {
// 到红黑树中查找
if (p instanceof TreeNode)
node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
// 在链表中查找
else {
do {
if (e.hash == hash &&
((k = e.key) == key ||
(key != null && key.equals(k)))) {
node = e;
break;
}
p = e;
} while ((e = e.next) != null);
}
}
// 删除node结点
if (node != null && (!matchValue || (v = node.value) == value ||
(value != null && value.equals(v)))) {
// 红黑树情况，
if (node instanceof TreeNode)
((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
// node=p，第一个结点就是要删除的结点，直接tab[index] = node.next;
else if (node == p)
tab[index] = node.next;
else
p.next = node.next;
++modCount;
--size;
// 这个方法是一个空方法，留给子类覆写，如LinkedHashMap就覆写了这个方法
afterNodeRemoval(node);
return node;
}
}
return null;
}
/**
* Removes all of the mappings from this map.
* The map will be empty after this call returns.
* 将tab置null
*/
public void clear() {
Node<K,V>[] tab;
modCount++;
if ((tab = table) != null && size > 0) {
size = 0;
for (int i = 0; i < tab.length; ++i)
tab[i] = null;
}
}
/**
* Returns <tt>true</tt> if this map maps one or more keys to the
* specified value.
*
* @param value value whose presence in this map is to be tested
* @return <tt>true</tt> if this map maps one or more keys to the
* specified value
* 挨个遍历判断
*/
public boolean containsValue(Object value) {
Node<K,V>[] tab; V v;
if ((tab = table) != null && size > 0) {
for (int i = 0; i < tab.length; ++i) {
for (Node<K,V> e = tab[i]; e != null; e = e.next) {
if ((v = e.value) == value ||
(value != null && value.equals(v)))
return true;
}
}
}
return false;
}
}/**
* Returns a power of two size for the given target capacity.
* 返回大于输入参数且最近的2的整数次幂的数。如cap=10,返回结果就是16
*/
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;
}先来分析有关n位操作部分：先来假设n的二进制为01xxx...xxx。接着
对n右移1位：001xx...xxx，再位或：011xx...xxx
对n右移2为：00011...xxx，再位或：01111...xxx
此时前面已经有四个1了，再右移4位且位或可得8个1
同理，有8个1，右移8位肯定会让后八位也为1。
综上可得，该算法让最高位的1后面的位全变为1。
最后再让结果n+1，即得到了2的整数次幂的值了。
现在回来看看第一条语句：
int n = cap - 1;让cap-1再赋值给n的目的是另找到的目标值大于或等于原值。例如二进制1000，十进制数值为8。如果不对它减1而直接操作，将得到答案10000，即16。显然不是结果。减1后二进制为111，再进行操作则会得到原来的数值1000，即8。
- 当数组长度为2的幂次方时，可以使用位运算来计算元素在数组中的下标
- 增加hash值的随机性，减少hash冲突
看一下hashmap的源码：
//计算hash值
static final int hash(Object key) {
int h;
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}h >>> 16表示无符号右移16位，低位挤走，高位补0；^ 为按位异或，即转成二进制后，相异为1，相同为0；由此可发现，当传入的值小于 2的16次方-1 时，调用这个方法返回的值，都是自身的值。
右位移16位，正好是32bit的一半，自己的高半区和低半区做异或，就是为了混合原始哈希码的高位和低位，以此来加大低位的随机性。而且混合后的低位掺杂了高位的部分特征，这样高位的信息也被变相保留下来。 假如没有进行高位运算，那最后参与运算的永远只是取模运算的最后几位，相似性会比较大。
hash函数的主要作用就是：增大随机性，减少碰撞。
hashmap中put的源码:
public V put(K key, V value) {
return putVal(hash(key), key, 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
* 核心方法之一，存入key、value
*/
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
// 如果table=null或者tab.length=0，就进行扩容
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
// 此处对应没有产生hash冲突的情况，直接使tab[i]等于一个新创建的Node结点
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else { // 产生hash冲突
Node<K,V> e; K k;
// 第一个结点就是要找的结点
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;
else if (p instanceof TreeNode)
// 到红黑树中查找
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
// p.next = null表示没有找到，直接插入一个新结点
p.next = newNode(hash, key, value, null);
// 如果binCount大于等于7，就转化为红黑树
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;
}
}
// 更新结点值
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
// 如果大于阈值，就进行扩容
if (++size > threshold)
resize();
afterNodeInsertion(evict);
return null;
}里面非常巧妙的代码是p = tab[i = (n - 1) & hash]，tab是hashmap存放存放元素的数组，(n - 1) & hash也解释了为什么table的大小要是2的倍数。
如果n是默认大小16，没有扩容，(n - 1) & hash的计算结果就是hash值本身；见下图：
如果n扩容，扩容大小是原大小*2，为什么n一定要是2的倍数？举个例子：
设：oldCap=16 二进制为：0001 0000 // oldCap是扩容之前的table大小
oldCap-1=15 二进制为：0000 1111 // 如果oldCap是2的倍数，低位就全部是1，与hash进行&运算，hash值不变
e1.hash=10 二进制为：0000 1010
e2.hash=26 二进制为：0101 1010
e1在扩容前的位置为：e1.hash & oldCap-1 结果为：0000 1010
e2在扩容前的位置为：e2.hash & oldCap-1 结果为：0000 1010
结果相同，所以e1和e2在扩容前在同一个链表上，这是扩容之前的状态。
现在扩容后，需要重新计算元素的位置，在扩容前的链表中计算地址的方式为e.hash & oldCap-1
那么在扩容后应该也这么计算呀，扩容后的容量为oldCap*2=32 0010 0000 newCap=32，新的计算
方式应该为
e1.hash & newCap-1
即：0000 1010 & 0001 1111
结果为0000 1010与扩容前的位置完全一样。
e2.hash & newCap-1
即：0101 1010 & 0001 1111
结果为0001 1010,为扩容前位置+oldCap。由此可见，扩容后的hashmap本来存储的元素，元素的hash值比原容量小，扩容后位置不变，元素的hash值比原容量大，扩容后的位置就是扩容前位置+原容量，直接可以计算出扩容后的位置，减少了一次求hash值的次数(不需要像JDK1.7的实现那样重新计算hash)；而且将有冲突的数据均匀的分散到新的空间上；而且&运算比%取模运算要快；
当桶上的结点数大于8会转成红黑树：
红黑树的查找速度更快，查找速度优化为O(logn)。
红黑树知识可以查看我的另一篇教程，红黑树。或者下面的教程
HashMap每次扩容都是建立一个新的table数组，长度和容量阈值都变为原来的两倍，然后把原数组元素重新映射到新数组上，具体步骤如下：
- 首先会判断table数组长度，如果大于0说明已被初始化过，那么
按当前table数组长度的2倍进行扩容，阈值也变为原来的2倍 - 若table数组未被初始化过，且threshold(阈值)大于0说明调用了
HashMap(initialCapacity, loadFactor)构造方法，那么就把数组大小设为threshold - 若table数组未被初始化，且threshold为0说明调用
HashMap()构造方法，那么就把数组大小设为16，threshold设为16*0.75 - 接着需要判断如果不是第一次初始化，那么扩容之后，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，如果节点是红黑树类型的话则需要进行红黑树的拆分。
这里有一个需要注意的点就是在JDK1.8 HashMap扩容阶段重新映射元素时不需要像1.7版本那样重新去一个个计算元素的hash值，而是通过hash & oldCap的值来判断，若为0则索引位置不变，不为0则新索引=原索引+旧数组长度，为什么呢？具体原因如下：
因为我们使用的是2次幂的扩展(指长度扩为原来2倍)，所以，元素的位置要么是在原位置，要么是在原位置再移动2次幂的位置。因此，我们在扩充HashMap的时候，不需要像JDK1.7的实现那样重新计算hash，只需要看看原来的hash值新增的那个bit是1还是0就好了，是0的话索引没变，是1的话索引变成“原索引+oldCap
为什么用hash & oldCap做判断，假设oldCap=16，二进制是10000，15的二进制是1111，hash & (length - 1)用了四位二进制可以得到所在的索引，如果需要扩容，就是原容量*2，也就是取5位二进制与hash做与算法，这里用hash & oldCap这种非常巧妙的方法，判断hash的倒数第五位二进制是不是1，如果是1，说明应该索引变成“原索引+oldCap”，为什么不用重新hash?因为重新hash的结果也是原索引+oldCap，无非就是用五位二进制算一下，通过hash & oldCap这种方式已经判断了；如果是hash & oldCap= 0，说明原位置不用动。这里非常巧妙，如果还不懂，我也没办法了，你们仔细想想。
扩容代码：
final Node<K,V>[] resize() {
// 旧数组
Node<K,V>[] oldTab = table;
// 旧容量
int oldCap = (oldTab == null) ? 0 : oldTab.length;
// 旧阈值
int oldThr = threshold;
// 新容量，新阈值
int newCap, newThr = 0;
if (oldCap > 0) {
// 旧容量大于等于最大容量，则阈值直接等于int最大值，返回原数组，无法进行扩容了
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
// 新容量设置为旧容量的2倍，新阈值也设置为旧阈值的2倍
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
// 新容量设置为旧阈值
newCap = oldThr;
else { // zero initial threshold signifies using defaults
// 旧容量等于0的情况，说明刚进行插入，使用默认容量
newCap = DEFAULT_INITIAL_CAPACITY;
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
}
// 设置新容量的值
if (newThr == 0) {
float ft = (float)newCap * loadFactor;
newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
(int)ft : Integer.MAX_VALUE);
}
threshold = newThr;
// 扩容就是新建一个数组，把原数组里面的数据存入新数组
@SuppressWarnings({"rawtypes","unchecked"})
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
if (oldTab != null) {
// 下面这个循环，就是将原数组中的数据存到新数组
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
if ((e = oldTab[j]) != null) {
oldTab[j] = null;
// 数组j的位置只有一个元素
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
else if (e instanceof TreeNode)
// 对应红黑树的情况
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
// 链表
// loHead和loTail用于记录oldTab[j]中结点，扩容后索引位置不变的情况
Node<K,V> loHead = null, loTail = null;
// hiHead和hiTail用于记录oldTab[j]中结点，扩容后索引位置等于原位置+原容量的情况
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
do {
next = e.next;
// 这个if成立，说明e结点扩容后还是在j位置
if ((e.hash & oldCap) == 0) {
// 尾插法
if (loTail == null)
loHead = e;
else
loTail.next = e;
loTail = e;
}
else {
// 这个if成立，说明e结点扩容后，存储在j+oldCap位置，也是尾插法
if (hiTail == null)
hiHead = e;
else
hiTail.next = e;
hiTail = e;
}
} while ((e = next) != null);
// loHead链表扩容后还在j位置
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
// hiHead链表对应新数组的位置就是，j + oldCap，这里也就是为什么数组大小一定要是2的倍数
// 注意到，这里不需要重新索引位置，在原容量 & hash=1的情况下，如果重新计算(原容量*2 - 1) & hash，
// 其计算结果也是j + oldCap, 这就是hashmap非常巧妙的地方。
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}因为容器中节点分布在hash桶中的频率遵循lambda=0.5的泊松分布，桶的长度超过8的概率非常非常小，约0.00000006。所以一般情况下都不会转为红黑树，如果你自己写的类当做hashmap的key，实现了hashcode和equals方法，hashcode写的太烂，就有可能导致hash桶中元素超过8，避免查找、删除效率太低，所以要转为红黑树。
因为红黑树占用空间是链表的两倍，而且当链表长度短时，红黑树不一定比链表快。
多线程环境下，HashMap 1.8依然会出现死循环的情况，发生在向红黑树添加节点中。多跑几遍下面代码可以跑出死循环。
import java.util.HashMap;
import java.util.Map;
import java.util.concurrent.atomic.AtomicInteger;
/**
* jdk7 扩容时都可能导致死锁
* jdk8 在PutTreeValue时可能死循环 死循环在hashMap的1816行或2229行， java version "1.8.0_111"
* jstack发现可能卡在 at java.util.HashMap$TreeNode.balanceInsertion(HashMap.java:2229)
* 也有可能卡在 at java.util.HashMap$TreeNode.root(HashMap.java:1816)
*
* @since 2019-02-23
*/
public class HashMap1 {
public static void main(String[] args) {
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(1);
for (int i = 0; i < 200; i++) {
new HashMapThread(map).start();
}
}
}
class HashMapThread extends Thread {
private static AtomicInteger ai = new AtomicInteger(0);
private HashMap<Integer, Integer> map;
HashMapThread(HashMap<Integer, Integer> map) {
this.map = map;
}
@Override
public void run() {
while (ai.get() < 100000) {
map.put(ai.get(), ai.get());
ai.incrementAndGet();
}
System.out.println(Thread.currentThread().getName() + "执行结束完");
}
}- https://www.jianshu.com/p/dc13f825b71f
- https://www.cnblogs.com/duodushuduokanbao/p/9492952.html
- https://blog.csdn.net/qq_37113604/article/details/81353626
- https://www.zhihu.com/question/20733617/answer/111577937
欢迎关注我的公众号呦，率先更新内容，并且后续还有一些源码级的免费教程推出。
