JDK源码学习之HashMap
HashMap源码学习
- HashMap类图
- HashMap概述
- HashMap原理
- HashMap中实现的细节
HashMap类图
HashMap概述
- 允许null key和null value
- HashMap与HashTable紧密相关，除了不是同步的，以及允许null值外。
- HashMap对迭代元素的顺序不保证，特别是，不保证迭代的顺序一致不变。
- 假定Hash函数分散的元素比较合理，它会提供常数时间的get 、put操作
- 如果迭代性能很重要，不把初始容量设置太高或者加载因子太低是很重要的。更高的值，会降低空间开销，但是会增加查找成本的开销。因此，当设定HashMap的初始容量时，应该考虑map集合中entry的预期数量和它的加载因子，以便减少rehash操作的次数。
HashMap原理
HashMap中实现的细节
- 在创建HashMap的时候，仅仅是对初始容量和加载因子做了赋值操作，并没有初始化数组操作，而是将数组的创建放到了第一次put中去。（用到才创建空间）
- transient修饰了table，entrySet，size，modCount。
Q：为什么用transient修饰这四个成员变量？
A： HashMap是经常会被序列化用于数据传输的。因此，序列化传输数据，我们关心的是接收到的数据，而不关新他的结构等。那么，当对方接受这些数据时，要怎么从数据中重新构建结构？HashMap手动实现了序列化反序列化方法writeObject readObject。
**注意：**用transient修饰的成员变量，会包含两层含义：
第一层：就是该变量确实不需要序列化传输，比如数据结构。
第二层：在明确需要手动序列化反序列化对象时，它的成员变量都要transient。 - HashMap中的容量先做减1再运算。对任意一个输入的容量，减1的操作是为了便于进行二进制运算。
/**
* 给定任意一个输入的容量，返回是2的整数权重大小的值。
* 为什么选择1 2 4 8 16，因为cap是int类型，为32位，所以最高移位就16位
* 1 2 4 8 移的都是2的倍数位，为了使得HashMap的容量都是2的幂。
* 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;
}
- 计算Hash值，也是用的二进制运算。
/**
* 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;
// 移16位，是为了避免高位的影响
return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
- 为了解决某个bin的链表过长，HashMap也采用了红黑树
/**
* 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.
* 触发树化的条件，但不一定就树化了，还需要满足table的length为64
*/
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.
*/
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.
* 如果，table的length少于64，即便某个bin中链表长度达到TREEIFY_THRESHOLD，依然不会树化，而是触发扩容
*/
static final int MIN_TREEIFY_CAPACITY = 64;
事例：
// 在容量为16的HashMap中，这些值都被存储在下标为3的bucket中。
// 当163被放到集合中的时候，会触发树化条件TREEIFY_THRESHOLD
// 同时，又判断当前HashMap的容量是否小于MIN_TREEIFY_CAPACITY
// 如果，小于，触发HashMap的扩容操作
// 大于，则进行下标为3的bucket的将链表进行树化
int[] data = new int[] {19，35, 51, 67, 83, 99, 115, 131, 147};
Map<Integer, Integer> t1 = new HashMap<>();
for (int i : data) {
t1.put(i, i);
}
Q：HashMap是怎么通过扩容来解决某个bin的链表长度过长这种情形的？
A：将bucket中，链长度超过阈值8的bin在非树化的情况下，扩容。以默认容量16为例说明。
经过Hash计算【index = 16 % data[i]】，data数组中的值存储到HashMap中，都将会落到下标为3的bin中去。当下标为3的bin的链表长度为8，且又不满足HashMap容量至少为64，触发HashMap的扩容resize操作。
比如，原来只有一个bin的链表过长，经过扩容，会将该链表上的这些元素一分为2。
比如原来为扩容之前的HashMap部分结构：
bucket idx table[] list capacity = 16
3 19 -> 35 -> 51 -> 67 -> 83 -> 99 -> 115 -> 131 -> 147
当向HashMap集合中存放147元素时，触发一倍的扩容 capacity = 32
bucket idx table[] list capacity = 32
3 35 -> 67 -> 99 -> 131
19 19 -> 51 -> 83 -> 115 -> 147
从这方面来说，这也是HashMap为什么每次扩容都扩一倍。 为了解决Hash碰撞 。从存放在3变成到3和19这个过程是resize的preserve order的过程。
/**
* 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;
// 扩容的
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;
// 比如，原来的数组中存放在table中的bin个数超过8，意味着这个8个元素的Hash计算后都落到同一个下标中
// 经过一倍扩容后，原来的8个Hash冲突的在同一个下标中，现在会被重新计算，分配到两个下标（按照冲突的下标个数，
// 比如，原来只有一个下标冲突，那么新扩容后用两个下标存放冲突值；如果原来有两个下标都有冲突，那么就会有四个下标
// 重新计算Hash值存放）中。
newTab[j] = loHead;
}
if (hiTail != null) {
hiTail.next = null;
// 将Hash值冲突的存放到扩容一倍后的数组中。
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}
- HashMap是怎么树化的？
还以data中的数据为例，当put147的时候，触发了HashMap的第一次扩容16->32。要想达到树化，还需进行第二次扩容由32->64的过程。因此，此时的data数组中的数据元素应该为：
int[] data = new int[] {19, 35, 51, 67, 83, 99, 115, 131, 147, 163, 179, 195, 211, 227, 243, 259, 275, 291, 307, 323, 339, 355, 371, 387, 403, 419, 435, 451, 467, 483, 499, 515, 531}当向HashMap中添加275元素触发该次扩容,继续添加后续元素至531触发树化
bucket idx table[] list capacity = 64
3 67 -> 131 -> 195 -> 259 -> 323 -> 387 -> 451 -> 515
19 19 -> 83 -> 147 -> 211 -> 275 -> 339 -> 403 -> 467 -> 531
35 35 -> 99 -> 163 -> 227 -> 291 -> 355 -> 419 -> 483
51 51 -> 115 -> 179 -> 243 -> 307 -> 371 -> 435 -> 499
/**
* Forms tree of the nodes linked from this node.
*/
final void treeify(Node<K,V>[] tab) {
TreeNode<K,V> root = null;
for (TreeNode<K,V> x = this, next; x != null; x = next) {
next = (TreeNode<K,V>)x.next;
x.left = x.right = null;
if (root == null) {
x.parent = null;
x.red = false;
root = x;
}
else {
K k = x.key;
int h = x.hash;
Class<?> kc = null;
for (TreeNode<K,V> p = root;;) {
int dir, ph;
K pk = p.key;
if ((ph = p.hash) > h)
dir = -1;
else if (ph < h)
dir = 1;
else if ((kc == null &&
(kc = comparableClassFor(k)) == null) ||
(dir = compareComparables(kc, k, pk)) == 0)
dir = tieBreakOrder(k, pk);
TreeNode<K,V> xp = p;
if ((p = (dir <= 0) ? p.left : p.right) == null) {
// 这些做的是，先向树中添加Node结点，然后再满足红黑树性质
x.parent = xp;
if (dir <= 0)
xp.left = x;
else
xp.right = x;
root = balanceInsertion(root, x);
break;
}
}
}
}
// 确保树化后的根节点在tab数组中
moveRootToFront(tab, root);
}