扩容部分我们需要重点关注,其余内容可以阅读上篇博客
扩容原理
1.扩容需要将桶的长度,更新的新长度;
2.将所有对象按照新的桶长度运算对应桶下标,
步骤2示例:
为了简单易于理解此处使用取余数代替源码中的(n - 1) & hash
扩容前,桶长度为2,分布如下:
table[0] 4,8
table[1] 3 ,6
扩容后,桶长度为4,分布如下:
table[0] 4,8
table[1] null
table[2] 6
table[3] 3
代码走读
理解完大致原理,我们来看下JDK8源代码
final Node<K,V>[] resize() {
Node<K,V>[] oldTab = table;
int oldCap = (oldTab == null) ? 0 : oldTab.length;
//threshold用于标记达到扩张的值正常情况为容量*负载因子
int oldThr = threshold;
int newCap, newThr = 0;
//计算新的容量值
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 翻倍
}
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
else { // zero initial threshold signifies using defaults 初始化
newCap = DEFAULT_INITIAL_CAPACITY;
//threshold初始化
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"})
//分配新的桶对象
Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
table = newTab;
if (oldTab != null) {
//循环按照新的容量进行重新存储
for (int j = 0; j < oldCap; ++j) {
Node<K,V> e;
//第一个节点不为空 说明存在对应链表或者二叉树
if ((e = oldTab[j]) != null) {
oldTab[j] = null;
//只有一个节点
if (e.next == null)
newTab[e.hash & (newCap - 1)] = e;
//二叉树的情况
else if (e instanceof TreeNode)
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order
//链表的情况 逐个遍历
Node<K,V> loHead = null, loTail = null;
Node<K,V> hiHead = null, hiTail = null;
Node<K,V> next;
//将链表按照e.hash & oldCap分成两个链表
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);
//loHead 保持在新的桶内的相同位置
if (loTail != null) {
loTail.next = null;
newTab[j] = loHead;
}
//hiHead 放置在新的桶内的j+oldCap位置
if (hiTail != null) {
hiTail.next = null;
newTab[j + oldCap] = hiHead;
}
}
}
}
}
return newTab;
}










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