1.1 ConcurrentHashMap put方法介绍
2.1 put方法解读
调用put(K key, V value)方法实际调用的是putVal(key, value, false),如下:
public V put(K key, V value) {
return putVal(key, value, false);
}
所以,解读重点在putVal方法上。先看一下putVal方法操作主要流程:
1.若 key == null || value == null,抛出 NPE,结束;否则,计算key的hash值,继续往下执行;
2.对存放
执行体进行如下一系列判断:
2.1 判断tab是否初始化,未初始化则进行初始化(首次初始化默认数组长度为16);
if (tab == null || (n = tab.length) == 0)
tab = initTable();
2.2 判断key对应的数组位置上是否为null,若尚未发生hash碰撞,即进行CAS操作,new 一个 Node
else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {
if (casTabAt(tab, i, null,
new Node(hash, key, value, null)))
break; // no lock when adding to empty bin
}
2.3 判断tab是否需要进行扩容:
else if ((fh = f.hash) == MOVED) // MOVED = -1
tab = helpTransfer(tab, f);
2.4 以上判断都未进入,说明tab已初始化,且有容量存放数据,但发生了hash冲突,接下来则进入hash冲突解决阶段:
先用synchronized 锁住f,即tab数组发生hash冲突位置上的元素对象Node,然后再判断一下tabAt(tab, i) == f,主要作用是tab是一个volatile修饰的对所有线程可见的共享变量,在synchronized 上锁成功之前其他线程有可能对tab做了操作,如扩容,remove等。
2.4.1 if (fh >= 0) 当做链表处理:进行for循环,若hash 和 key 都相同,则用新值替换旧值,break;否则一直循环到链表尾,当 (e = e.next) == null 时,pred.next = new Node
2.4.2 else if (f instanceof TreeBin) 当做树结构处理:若添加的key不存在于树中,则 putTreeVal 方法直接添加并返回null,若是key对应的数据已存在,putTreeVal 方法返回树中的Node
2.4.3 在synchronized 代码块执行结束后,判断链表长度是否 >= 阈值 8(TREEIFY_THRESHOLD),是则转为红黑树;
之后要么 return oldVal ,要么break 结束for循环;
else {
V oldVal = null;
synchronized (f) {
if (tabAt(tab, i) == f) {
if (fh >= 0) {
binCount = 1;
for (Node e = f;; ++binCount) {
K ek;
if (e.hash == hash &&
((ek = e.key) == key ||
(ek != null && key.equals(ek)))) { // key 存在,更新
oldVal = e.val;
if (!onlyIfAbsent)
e.val = value;
break;
}
Node pred = e;
if ((e = e.next) == null) {
pred.next = new Node(hash, key,
value, null); //key 不存在,链表中追加新元素
break;
}
}
}
else if (f instanceof TreeBin) {
Node p;
binCount = 2;
//key不存在则putTreeVal方法直接添加新元素并返回null,key存在则返回对应节点p并做val更新
if ((p = ((TreeBin)f).putTreeVal(hash, key,
value)) != null) {
oldVal = p.val;
if (!onlyIfAbsent)
p.val = value;
}
}
}
}
if (binCount != 0) {
// TREEIFY_THRESHOLD 默认8 链表转红黑树
if (binCount >= TREEIFY_THRESHOLD)
treeifyBin(tab, i);
if (oldVal != null)
return oldVal;
break;
}
}
3. for 循环处理结束后,检查tab是否需要扩容:
addCount(1L, binCount); //检查扩容
4. putVal 方法执行结束: return null;
2.2 put 方法流程图
为更直观了解put方法的执行过程,画个流程图:
2.3 细节探讨
2.3.1 链表转红黑树
1. 链表转红黑树阈值 TREEIFY_THRESHOLD 8 的问题,看源码,是用binCount来比较;但binCount并不能完全正确的及时反映链表的长度。看下面这段代码:
synchronized (f) {
if (tabAt(tab, i) == f) {
if (fh >= 0) {
binCount = 1;
for (Node e = f;; ++binCount) {
K ek;
if (e.hash == hash &&
((ek = e.key) == key ||
(ek != null && key.equals(ek)))) {
oldVal = e.val;
if (!onlyIfAbsent)
e.val = value;
break;
}
Node pred = e;
if ((e = e.next) == null) {
pred.next = new Node(hash, key,
value, null);
break;
}
}
}
else if (f instanceof TreeBin) {
Node p;
binCount = 2;
if ((p = ((TreeBin)f).putTreeVal(hash, key,
value)) != null) {
oldVal = p.val;
if (!onlyIfAbsent)
p.val = value;
}
}
}
}
if (binCount != 0) {
if (binCount >= TREEIFY_THRESHOLD)
treeifyBin(tab, i);
if (oldVal != null)
return oldVal;
break;
}
(1)假设本次key存在,且在链表的中间位置,那么break的时候 binCount 不能代表链表的实际长度,此时实际长度可能已达 到8;
(2)假设本次key不存在,且链表长度已经为7,那么添加新元素后,实际链表长度已经是8,但break时,binCount 还是7;
(3)至于阈值为什么为8,哈哈,不清楚哈,估计是基于测试统计给出来的吧。有个 泊松分布 概率函数,感兴趣的可以了解 下。
2. 链表转红黑树,关于tab 数组长度的问题。看代码:
/**
* Replaces all linked nodes in bin at given index unless table is
* too small, in which case resizes instead.
*/
private final void treeifyBin(Node[] tab, int index) {
Node b; int n, sc;
if (tab != null) {
if ((n = tab.length) < MIN_TREEIFY_CAPACITY)
tryPresize(n << 1); //MIN_TREEIFY_CAPACITY 默认64,tryPresize 扩容
else if ((b = tabAt(tab, index)) != null && b.hash >= 0) {
synchronized (b) {
if (tabAt(tab, index) == b) {
//红黑树转化
TreeNode hd = null, tl = null;
for (Node e = b; e != null; e = e.next) {
TreeNode p =
new TreeNode(e.hash, e.key, e.val,
null, null);
if ((p.prev = tl) == null)
hd = p;
else
tl.next = p;
tl = p;
}
setTabAt(tab, index, new TreeBin(hd));
}
}
}
}
}
(1)MIN_TREEIFY_CAPACITY 为64,可见,当链表长度达到8但tab长度小于64时,并不会直接进行红黑树转化,而是对tab进行扩容,根据新的new_tab大小以及key的hash值,重新分配index,将原有元素copy到新的new_tab中,且原有元素若为链表,扩容后理论上分为2个小链表,链表位置为 index=i 和 index=i+n ,n 表示原有tab大小;若原有元素为树结构,扩容后理论上分为2个小树,树位置为 index=i 和 index=i+n ,n 表示原有tab大小,若小树中元素小于等于UNTREEIFY_THRESHOLD 6 ,扩容时会向下转化为链表,否则继续生成树结构;由此也可见,当链表长度达到8但tab长度小于64时,采用的是扩容来降低hash冲突,从而理论上降低链表长度。
3.1 ConcurrentHashMap 1.8 put 方法主要采用 CAS (hash未冲突使用) + synchronized (hash冲突使用),来解决并发时的线程安全问题;
3.2 链表转为红黑树,必要条件是 数组长度不小于64,且链表长度不小于8;
/**
* Maps the specified key to the specified value in this table.
* Neither the key nor the value can be null.
*
* The value can be retrieved by calling the {@code get} method
* with a key that is equal to the original key.
*
* @param key key with which the specified value is to be associated
* @param value value to be associated with the specified key
* @return the previous value associated with {@code key}, or
* {@code null} if there was no mapping for {@code key}
* @throws NullPointerException if the specified key or value is null
*/
public V put(K key, V value) {
return putVal(key, value, false);
}
/** Implementation for put and putIfAbsent */
final V putVal(K key, V value, boolean onlyIfAbsent) {
if (key == null || value == null) throw new NullPointerException();
int hash = spread(key.hashCode());
int binCount = 0;
for (Node[] tab = table;;) {
Node f; int n, i, fh;
if (tab == null || (n = tab.length) == 0)
tab = initTable();
else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {
if (casTabAt(tab, i, null,
new Node(hash, key, value, null)))
break; // no lock when adding to empty bin
}
else if ((fh = f.hash) == MOVED)
tab = helpTransfer(tab, f);
else {
V oldVal = null;
synchronized (f) {
if (tabAt(tab, i) == f) {
if (fh >= 0) {
binCount = 1;
for (Node e = f;; ++binCount) {
K ek;
if (e.hash == hash &&
((ek = e.key) == key ||
(ek != null && key.equals(ek)))) {
oldVal = e.val;
if (!onlyIfAbsent)
e.val = value;
break;
}
Node pred = e;
if ((e = e.next) == null) {
pred.next = new Node(hash, key,
value, null);
break;
}
}
}
else if (f instanceof TreeBin) {
Node p;
binCount = 2;
if ((p = ((TreeBin)f).putTreeVal(hash, key,
value)) != null) {
oldVal = p.val;
if (!onlyIfAbsent)
p.val = value;
}
}
}
}
if (binCount != 0) {
if (binCount >= TREEIFY_THRESHOLD)
treeifyBin(tab, i);
if (oldVal != null)
return oldVal;
break;
}
}
}
addCount(1L, binCount);
return null;
}
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