当我们享受着jdk带来的便利时同样承受它带来的不幸恶果。通过分析Hashtable就知道，synchronized是针对整张Hash表的，即每次锁住整张表让线程独占，安全的背后是巨大的浪费，而现在的解决方案----ConcurrentHashMap。
public V get(Object key) {
Node<K,V>[] tab; Node<K,V> e, p; int n, eh; K ek;
int h = spread(key.hashCode());
if ((tab = table) != null && (n = tab.length) > 0 &&
(e = tabAt(tab, (n - 1) & h)) != null) {
if ((eh = e.hash) == h) {
if ((ek = e.key) == key || (ek != null && key.equals(ek)))
return e.val;
}
else if (eh < 0)
return (p = e.find(h, key)) != null ? p.val : null;
while ((e = e.next) != null) {
if (e.hash == h &&
((ek = e.key) == key || (ek != null && key.equals(ek))))
return e.val;
}
}
return null;
}
put操作一上来就锁定了整个segment，这当然是为了并发的安全，修改数据是不能并发进行的，必须得有个判断是否超限的语句以确保容量不足时能够rehash，，原来segment里面才是真正的hashtable，即每个segment是一个传统意义上的hashtable,如上图，从两者的结构就可以看出区别，这里就是找出需要的entry在table的哪一个位置，之后得到的entry就是这个链的第一个节点，如果e!=null，说明找到了，这是就要替换节点的值if(!onlyIfAbsent)，否则，我们需要new一个entry，它的后继是first，而让tab[index]指向它，什么意思呢？实际上就是将这个新entry插入到链头，剩下的就非常容易理解了。
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;
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<K,V>(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<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;
}
remove操作非常类似put，但要注意一点区别，中间那个for循环是做什么用的呢？从代码来看，就是将定位之后的所有entry克隆并拼回前面去，但有必要吗？每次删除一个元素就要将那之前的元素克隆一遍？这点其实是由entry的不变性来决定的，仔细观察entry定义，发现除了value，其他所有属性都是用final来修饰的，这意味着在第一次设置了next域之后便不能再改变它，取而代之的是将它之前的节点全都克隆一次。
public V remove(Object key) {
return replaceNode(key, null, null);
}
/**
* Implementation for the four public remove/replace methods:
* Replaces node value with v, conditional upon match of cv if
* non-null. If resulting value is null, delete.
*/
final V replaceNode(Object key, V value, Object cv) {
int hash = spread(key.hashCode());
for (Node<K,V>[] tab = table;;) {
Node<K,V> f; int n, i, fh;
if (tab == null || (n = tab.length) == 0 ||
(f = tabAt(tab, i = (n - 1) & hash)) == null)
break;
else if ((fh = f.hash) == MOVED)
tab = helpTransfer(tab, f);
else {
V oldVal = null;
boolean validated = false;
synchronized (f) {
if (tabAt(tab, i) == f) {
if (fh >= 0) {
validated = true;
for (Node<K,V> e = f, pred = null;;) {
K ek;
if (e.hash == hash &&
((ek = e.key) == key ||
(ek != null && key.equals(ek)))) {
V ev = e.val;
if (cv == null || cv == ev ||
(ev != null && cv.equals(ev))) {
oldVal = ev;
if (value != null)
e.val = value;
else if (pred != null)
pred.next = e.next;
else
setTabAt(tab, i, e.next);
}
break;
}
pred = e;
if ((e = e.next) == null)
break;
}
}
else if (f instanceof TreeBin) {
validated = true;
TreeBin<K,V> t = (TreeBin<K,V>)f;
TreeNode<K,V> r, p;
if ((r = t.root) != null &&
(p = r.findTreeNode(hash, key, null)) != null) {
V pv = p.val;
if (cv == null || cv == pv ||
(pv != null && cv.equals(pv))) {
oldVal = pv;
if (value != null)
p.val = value;
else if (t.removeTreeNode(p))
setTabAt(tab, i, untreeify(t.first));
}
}
}
}
}
if (validated) {
if (oldVal != null) {
if (value == null)
addCount(-1L, -1);
return oldVal;
}
break;
}
}
}
return null;
}
原文：http://www.cnblogs.com/android-blogs/p/5761430.html