/**
* 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.)
*/
//存储数据的数组，第一次使用才会初始化，总是2的倍数
transient Node<K,V>[] table;
/**
* The next size value at which to resize (capacity * load factor).
* 扩容后的最大size，翻译：临界值
*/
int threshold;
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
//说明 为什么通过(n - 1) & hash 获取index
/**
1.&运算更快
2.当 n 是2的次幂时， n - 1 通过 二进制表示即尾端一直都是以连续1的形式表示的。当(n - 1) 与 hash 做与运算时，会保留hash中 后 x 位的 1，这样就保证了索引值 不会超出数组长度。
3.同时当n为2次幂时，会满足一个公式：(n - 1) & hash = hash % n。
*/
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
boolean evict) {
Node<K,V>[] tab; Node<K,V> p; int n, i;
//如果没有初始化过，先通过resize初始化
if ((tab = table) == null || (n = tab.length) == 0)
n = (tab = resize()).length;
//如果计算出的index下为空，直接放进去
if ((p = tab[i = (n - 1) & hash]) == null)
tab[i] = newNode(hash, key, value, null);
else {
Node<K,V> e; K k;
//判断key和当前index上的node的key是否相同
//hash相等？（key==key？且key。equels（key））
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
e = p;//相等，则跳出（下面直接修改value）
else if (p instanceof TreeNode)
//加到红黑树上
e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
else {
//加到链表上
for (int binCount = 0; ; ++binCount) {
if ((e = p.next) == null) {
//尾插
p.next = newNode(hash, key, value, null);
if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st 添加之后长度到8 转换为树
treeifyBin(tab, hash);//这个方法里判断是否>64
break;
}
/判断每个节点和新key是否相等，是则替换
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
if (e != null) { // existing mapping for key
V oldValue = e.value;
if (!onlyIfAbsent || oldValue == null)
e.value = value;
//linkedhashMap中用到 move node to last
afterNodeAccess(e);
return oldValue;
}
}
++modCount;
//新的size和当前的长度比较
if (++size > threshold)
resize();
//linkedhashMap中用到 用来回调移除最早放入Map的对象
afterNodeInsertion(evict);
return null;
}
(e.hash & oldCap) == 0) 重点方法，扩容时快速定位下标
final Node<K,V>[] resize() {
Node<K,V>[] oldTab = table;//旧数组
int oldCap = (oldTab == null) ? 0 : oldTab.length;//旧数组长度
int oldThr = threshold;//旧临界值
int newCap, newThr = 0;
if (oldCap > 0) {
//如果数组已经最大了，直接返回
if (oldCap >= MAXIMUM_CAPACITY) {
threshold = Integer.MAX_VALUE;
return oldTab;
}
//原来的长度》16且新长度x2没到极限
else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
oldCap >= DEFAULT_INITIAL_CAPACITY)
newThr = oldThr << 1; // double threshold
}
//数组为空，阈值不为0，表示初始化过
else if (oldThr > 0) // initial capacity was placed in threshold
newCap = oldThr;
else { //初始化 数组长度16，阈值0.75*16
newCap = DEFAULT_INITIAL_CAPACITY;
newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
}
//如果新的阈值没有初始化，（上面的分支二，初始化一下）
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;//创建一个新的数组
if (oldTab != null) {
for (int j = 0; j < oldCap; ++j) {//遍历旧的
Node<K,V> e;
if ((e = oldTab[j]) != null) {
oldTab[j] = null;//清除旧节点
if (e.next == null)//如果只有一个元素，重新hash放到新数组
newTab[e.hash & (newCap - 1)] = e;
else if (e instanceof TreeNode)//树
((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
else { // preserve order 链表
//取每一个元素重新hash，如果(e.hash & oldCap) == 0
//放到新数组的原索引，否则放到新数组的原索引+旧数组长度的位置
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;
}
public V remove(Object key) {
Node<K,V> e;
return (e = removeNode(hash(key), key, null, false, true)) == null ?
null : e.value;
}
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;
//map不为空，存在该key才remove，否则直接返回null。
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;
//如果index上第一个元素就是，赋值给node，后面操作
if (p.hash == hash &&
((k = p.key) == key || (key != null && key.equals(k))))
node = p;
else if ((e = p.next) != null) {
//否则到链表或者树上找
if (p instanceof TreeNode)
node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
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);
}
}
//
if (node != null && (!matchValue || (v = node.value) == value ||
(value != null && value.equals(v)))) {
if (node instanceof TreeNode)
//删除树节点
((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
else if (node == p)
//直接删除
tab[index] = node.next;
else
//链表删除
p.next = node.next;
++modCount;
--size;
afterNodeRemoval(node);
return node;
}
}
return null;
}
