背景

上文JDK8中的HashMap源码写了HashMap,这次写ConcurrentHashMap

ConcurrentHashMap源码

/**
* Maps the specified key to the specified value in this table.
* Neither the key nor the value can be null.
*
* <p>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<K,V>[] tab = table;;) {
Node<K,V> f; int n, i, fh;
//tab为空,则初始化
if (tab == null || (n = tab.length) == 0)
tab = initTable();
else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) { //该槽为空,则尝试插入
if (casTabAt(tab, i, null,
new Node<K,V>(hash, key, value, null)))
break; // no lock when adding to empty bin
}
else if ((fh = f.hash) == MOVED) //正在移动,
tab = helpTransfer(tab, f);
else {
V oldVal = null;
synchronized (f) { //对该槽进行加锁
if (tabAt(tab, i) == f) {
if (fh >= 0) {
binCount = 1;
for (Node<K,V> 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<K,V> pred = e;
if ((e = e.next) == null) {
pred.next = new Node<K,V>(hash, key,
value, null);
break;
}
}
}
else if (f instanceof TreeBin) {
Node<K,V> p;
binCount = 2;
if ((p = ((TreeBin<K,V>)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;
}
/**
* Returns the value to which the specified key is mapped,
* or {@code null} if this map contains no mapping for the key.
*
* <p>More formally, if this map contains a mapping from a key
* {@code k} to a value {@code v} such that {@code key.equals(k)},
* then this method returns {@code v}; otherwise it returns
* {@code null}. (There can be at most one such mapping.)
*
* @throws NullPointerException if the specified key is null
*/
public V get(Object key) {
Node<K,V>[] tab; Node<K,V> e, p; int n, eh; K ek;
//获得hash值
int h = spread(key.hashCode());
//表非空,且该处不为空
if ((tab = table) != null && (n = tab.length) > 0 &&
(e = tabAt(tab, (n - 1) & h)) != null) {
if ((eh = e.hash) == h) { //判断第1个
if ((ek = e.key) == key || (ek != null && key.equals(ek)))
return e.val;
}
else if (eh < 0) //eh<0,找其他的
return (p = e.find(h, key)) != null ? p.val : null;
while ((e = e.next) != null) { //遍历
if (e.hash == h &&
((ek = e.key) == key || (ek != null && key.equals(ek))))
return e.val;
}
}
return null;
}

ConcurrentHashMap代码太多了,粘了好几次粘不上来。只粘几个方法吧。

阅后感

ConcurrentHashMap通过几个原子操作尽量减少加锁操作。

扩容部分没有看太明白,尤其时扩容时进行get操作。后续再继续学习。

05-11 10:54