之前一直没了解过HashMap的扩容机制,相关的方法是resize()方法,下面看下这个方法。
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;
}
上面的是源码,这个方法有点长,我们可以拆成两段来理解,
- 原数组扩容,每次扩大之前的2倍
- 扩容后,元素的迁移
HashMap在首次添加元素或元素个数达到阈值时,会进行调用resize方法进行扩容操作。下面这段代码还是比较好理解的
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;
}
//扩大两倍
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
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];
//指向扩容后的数组
table = newTab;
我们重点看元素迁移这段代码
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)
//(1)
newTab[e.hash & (newCap - 1)] = e;
else if (e instanceof TreeNode)
//(2)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
//(3)
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) {
//元素位置未变
loTail.next = null;
newTab[j] = loHead;
}
if (hiTail != null) {
//元素位置变化,数组索引位置+扩容前数组长度
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
我们主要看代码中标记1、2、3的部分,
(1)数组索引上只有一个元素时,重新按扩容后的长度进行哈希计算确定索引位置。
(2)数组索引上的元素是红黑树时,进行相关树的操作,这个我们今天先不看。
(3)重点看标记3的这部分代码,表示当前数组索引上有多个元素时,有两种处理,索引位置不变+索引位置变化。
如何判断元素索引位置不变,看源码中是通过(e.hash & oldCap) == 0 判断的,为什么呢?
HashMap是以2的倍数扩容的,两者二进制位高位差1,
- 16用二进制表示是 00010000
- 32用二进制表示是 00100000
HashMap是通过 (n - 1) & hash 确定索引位置,
- 15用二进制表示是 00001111
- 31用二进制表示是 00011111
现有元素key为14,17,我们分别计算下在数组长度为16和32下的索引位置,
14用二进制表示是 00001110,
- 数组长度为16,00001110 & 00001111 =1(14与15与运算),索引值为14 ,00001110 &00010000 = 0(14与16与运算)
- 数组长度为32,00001110 & 00011111=17(14与31与运算),索引值为14,00001110 & 00100000= 0(14与32与运算)
17用二进制表示是 00010001,
- 数组长度为16,00010001 & 00001111 =1(17与15与运算) ,索引值为1 ,00010001 & 00010000 = 1(17与16与运算)
- 数组长度为32,00010001 & 00011111 =17(17与31与运算) ,索引值为17,00010001 & 00100000= 17(17与32与运算)
从以上可以看出, (e.hash & oldCap) == 0
- 为true时,元素索引位置不变。
- 为false时,元素索引位置为 数组索引位置+扩容前数组长度。