1.resize
public class HashMap<K,V> extends AbstractMap<K,V>implements Map<K,V>, Cloneable, Serializable {final Node<K,V>[] resize() {Node<K,V>[] oldTab = table;int oldCap = (oldTab == null) ? 0 : oldTab.length; //老的数组容量int oldThr = threshold;//老的阈值int newCap, newThr = 0;if (oldCap > 0) {//老的数组容量大于0，说明table初始化过了if (oldCap >= MAXIMUM_CAPACITY) { //已经达到最大容量threshold = Integer.MAX_VALUE;return oldTab;} else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&oldCap >= DEFAULT_INITIAL_CAPACITY){//新容量=老容量*2newThr = oldThr << 1; // double threshold}} else if (oldThr > 0){ // initial capacity was placed in thresholdnewCap = oldThr;//这里是大于初始化容量的最小的2的n次方} else { // zero initial threshold signifies using defaultsnewCap = DEFAULT_INITIAL_CAPACITY;//默认为16newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY); //16*0.75}if (newThr == 0) {float ft = (float)newCap * loadFactor;//赋值新阈值newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?(int)ft : Integer.MAX_VALUE);}threshold = newThr; //更新全局threshold@SuppressWarnings({"rawtypes","unchecked"})Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];//根据新容量创建新数组table = newTab;//更新全局tableif (oldTab != null) {//说明老数组有数据，需要分配到新数组for (int j = 0; j < oldCap; ++j) {//遍历数组Node<K,V> e;if ((e = oldTab[j]) != null) {//第j个桶oldTab[j] = null;//老数组的值赋值为null，方便回收if (e.next == null){//只有一个元素，新算出index，赋值给新数组newTab[index]newTab[e.hash & (newCap - 1)] = e;} else if (e instanceof TreeNode){//如果是树((TreeNode<K,V>)e).split(this, newTab, j, oldCap);} else { // preserve orderNode<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) {//满足这个条件的元素组成一个链表if (loTail == null){loHead = e;} else{loTail.next = e;}loTail = e;} else {//满足这个条件的元素组成一个链表if (hiTail == null){hiHead = e;} else{hiTail.next = e;}hiTail = e;}} while ((e = next) != null);if (loTail != null) {loTail.next = null;newTab[j] = loHead;//链表头}if (hiTail != null) {hiTail.next = null;newTab[j + oldCap] = hiHead;//链表头}}}}}return newTab;}
}2.流程图
3.(e.hash & oldCap) == 0的理解
&：按位与 两个数都为1的时候，才为1
(e.hash & oldCap) == 0 新索引：j（跟老索引一致）
(e.hash & oldCap) != 0 新索引：j + oldCap
3.1. (e.hash & oldCap) == 0
假设oldCap是16，二进制就是 0001 0000
(e.hash & oldCap) == 0 的时候，e.hash的从右到左第5位为0：xxx0 xxxx
oldIndex = e.hash & (oldCap -1)
xxx0 xxxx
0000 1111 15
===========
0000 xxxx
newIndex = e.hash & (2*oldCap - 1)
xxx0 xxxx
0001 1111 31
===========
0000 xxxx
所以，如果(e.hash & oldCap) == 0的时候，这些元素的新索引跟老索引一致。
3.2. (e.hash & oldCap) != 0
假设oldCap是16，二进制就是 0001 0000
(e.hash & oldCap) != 0 的时候，e.hash的从右到左第5位为1：xxx1 xxxx
oldIndex = e.hash & (oldCap -1)
xxx1 xxxx
0000 1111 15
===========
0000 xxxx
newIndex = e.hash & (2*oldCap - 1)
xxx1 xxxx
0001 1111 31
===========
0001 xxxx
所以，如果(e.hash & oldCap) != 0的时候，这些元素的新索引=老索引+oldCap。
4.疑惑
在挪动老的数据到新table的时候，为啥要这样计算呢？这样计算并没有省什么时间吧
计算在新table的位置e.hash & (newCap - 1)，这个跟e.hash & oldCap执行速度一样，为啥不直接用e.hash & (newCap - 1)计算?
if ((e.hash & oldCap) == 0) {//满足这个条件的元素组成一个链表if (loTail == null){loHead = e;} else{loTail.next = e;}loTail = e;
} else {//满足这个条件的元素组成一个链表if (hiTail == null){hiHead = e;} else{hiTail.next = e;}hiTail = e;
}
