1.首先从下面这段代码开始分析
ConcurrentHashMap<String, Object> map = new ConcurrentHashMap<>();
map.put("123", "1");
map.put("234", "2");
System.out.println(map.get("123"));
System.out.println(map.size());
2.我们先从put方法看起
public V put(K key, V value) {
return putVal(key, value, false);
}
final V putVal(K key, V value, boolean onlyIfAbsent) {
// 首先判断key和value都不能为null
if (key == null || value == null) throw new NullPointerException();
// 获取key对应的hash值
int hash = spread(key.hashCode());
int binCount = 0;
// 开始自旋遍历元素
for (Node<K,V>[] tab = table;;) {
Node<K,V> f; int n, i, fh;
// 如果数组为空，则进行初始化操作
if (tab == null || (n = tab.length) == 0)
tab = initTable();
// 先通过unsafe操作获取i位置的元素，并判断是否为空
else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {
// 通过cas操作在数组的i位置上创建一个Node元素
if (casTabAt(tab, i, null,
new Node<K,V>(hash, key, value, null)))
break; // no lock when adding to empty bin
}
// 如果当前key的hash值为-1,表示当前数组i位置上有线程正在扩容，那么这个时间就要帮忙要转移元素
else if ((fh = f.hash) == MOVED)
tab = helpTransfer(tab, f);
// 如果数组i位置上有元素，则走到下面代码逻辑
else {
V oldVal = null;
// 这里使用synchronized 对数组的i位置的f元素进行加锁
synchronized (f) {
// 再次判断第i位置上的元素是否还是等于f
if (tabAt(tab, i) == f) {
// 判断key的hash是否大于0
if (fh >= 0) {
binCount = 1;
// 链表下面插入元素逻辑，遍历链表
for (Node<K,V> e = f;; ++binCount) {
K ek;
// 如果遍历过程中key相等，则覆盖旧的值，并返回旧的
if (e.hash == hash &&
((ek = e.key) == key ||
(ek != null && key.equals(ek)))) {
oldVal = e.val;
if (!onlyIfAbsent)
e.val = value;
break;
}
// 如果遍历过程中没有找到，则插入到链表的尾部
Node<K,V> pred = e;
if ((e = e.next) == null) {
pred.next = new Node<K,V>(hash, key,
value, null);
break;
}
}
}
// 如果数组i位置的f元素是红黑树，则进行树插入操作，这里有比较和自旋操作
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;
}
}
}
}
// 如果f位置上链表长度大于8,则进行树化
if (binCount != 0) {
if (binCount >= TREEIFY_THRESHOLD)
treeifyBin(tab, i);
if (oldVal != null)
return oldVal;
break;
}
}
}
// 插入之后，对Map的容量进行加1操作
addCount(1L, binCount);
return null;
}
3.其中看一下上面的treeifyBin(tab, i);这个方法
private final void treeifyBin(Node<K,V>[] tab, int index) {
Node<K,V> b; int n, sc;
if (tab != null) {
if ((n = tab.length) < MIN_TREEIFY_CAPACITY)
tryPresize(n << 1);
else if ((b = tabAt(tab, index)) != null && b.hash >= 0) {
// 这里再次获取数组i位置的元素b,然后锁住这个对象
synchronized (b) {
// 再次判断i位置元素是否等于b
if (tabAt(tab, index) == b) {
TreeNode<K,V> hd = null, tl = null;
// 遍历链表
for (Node<K,V> e = b; e != null; e = e.next) {
TreeNode<K,V> p =
new TreeNode<K,V>(e.hash, e.key, e.val,
null, null);
if ((p.prev = tl) == null)
hd = p;
else
tl.next = p;
tl = p;
}
// 找到树的根节点hd，并进行树化new TreeBin<K,V>(hd)，其中TreeBin是包含红黑数根节点的一个对象
// 然后在数组i位置设置创建出来的new TreeBin<K,V>(hd)元素
setTabAt(tab, index, new TreeBin<K,V>(hd));
}
}
}
}
}
4.然后在看一下数组初始化的initTable方法
private final Node<K,V>[] initTable() {
Node<K,V>[] tab; int sc;
// 循环过程
while ((tab = table) == null || tab.length == 0) {
// 如果sc小于0,则进行放弃当前cpu操作，再去和其他线程进行cpu竞争
if ((sc = sizeCtl) < 0)
Thread.yield(); // lost initialization race; just spin
// 一个cas操作判断是否相等，并-1操作
else if (U.compareAndSwapInt(this, SIZECTL, sc, -1)) {
try {
// 进行数组初始化操作
if ((tab = table) == null || tab.length == 0) {
int n = (sc > 0) ? sc : DEFAULT_CAPACITY;
@SuppressWarnings("unchecked")
Node<K,V>[] nt = (Node<K,V>[])new Node<?,?>[n];
table = tab = nt;
sc = n - (n >>> 2);
}
} finally {
// 计算出阈值
sizeCtl = sc;
}
break;
}
}
return tab;
}
5.再看一下putVal方法中最后一段代码中的addCount方法
private final void addCount(long x, int check) {
CounterCell[] as; long b, s;
// 下面这个判断是，如果counterCells数组为null，那么就会走在数组的baseCount基础上去加1，如果失败则进入判断内
if ((as = counterCells) != null ||
!U.compareAndSwapLong(this, BASECOUNT, b = baseCount, s = b + x)) {
CounterCell a; long v; int m;
boolean uncontended = true;
// 下面一段代码的逻辑是对CounterCell数组，根据不同线程取一个随机数&数组长度，之后再数组对应位置赋值（在basecount的基础上+1）,如果存在则在之前的基础上加1，最后在遍历CounterCell数组算出每个数组value之和算出map的size
if (as == null || (m = as.length - 1) < 0 ||
(a = as[ThreadLocalRandom.getProbe() & m]) == null ||
!(uncontended =
U.compareAndSwapLong(a, CELLVALUE, v = a.value, v + x))) {
fullAddCount(x, uncontended);
return;
}
if (check <= 1)
return;
s = sumCount();
}
// 下面就是扩容情况
if (check >= 0) {
Node<K,V>[] tab, nt; int n, sc;
// 如果数组容量大于阈值，并且数组不为null，数组长度小于MAXIMUM_CAPACITY，并且会再次判断
while (s >= (long)(sc = sizeCtl) && (tab = table) != null &&
(n = tab.length) < MAXIMUM_CAPACITY) {
// 这里会计算出一个很大的负数
int rs = resizeStamp(n);
// 下面的判断是防止两个线程同时进行扩容
if (sc < 0) {
if ((sc >>> RESIZE_STAMP_SHIFT) != rs || sc == rs + 1 ||
sc == rs + MAX_RESIZERS || (nt = nextTable) == null ||
transferIndex <= 0)
break;
if (U.compareAndSwapInt(this, SIZECTL, sc, sc + 1))
transfer(tab, nt);
}
else if (U.compareAndSwapInt(this, SIZECTL, sc,
(rs << RESIZE_STAMP_SHIFT) + 2))
transfer(tab, null);
// 重新计算map的大小
s = sumCount();
}
}
}
6.再继续点击transfer方法查看扩容逻辑
private final void transfer(Node<K,V>[] tab, Node<K,V>[] nextTab) {
int n = tab.length, stride;
// 下面会计算出扩容的步长stride，这在多线程下计算的值使固定的
if ((stride = (NCPU > 1) ? (n >>> 3) / NCPU : n) < MIN_TRANSFER_STRIDE)
stride = MIN_TRANSFER_STRIDE; // subdivide range
// 初始化新数组，大小翻倍
if (nextTab == null) { // initiating
try {
@SuppressWarnings("unchecked")
Node<K,V>[] nt = (Node<K,V>[])new Node<?,?>[n << 1];
nextTab = nt;
} catch (Throwable ex) { // try to cope with OOME
sizeCtl = Integer.MAX_VALUE;
return;
}
nextTable = nextTab;
transferIndex = n;
}
int nextn = nextTab.length;
ForwardingNode<K,V> fwd = new ForwardingNode<K,V>(nextTab);
// 是否允许前进
boolean advance = true;
// 当前线程是否完成任务
boolean finishing = false; // to ensure sweep before committing nextTab
for (int i = 0, bound = 0;;) {
Node<K,V> f; int fh;
// 判断是否还可以继续前进
while (advance) {
int nextIndex, nextBound;
if (--i >= bound || finishing)
advance = false;
else if ((nextIndex = transferIndex) <= 0) {
i = -1;
advance = false;
}
else if (U.compareAndSwapInt
(this, TRANSFERINDEX, nextIndex,
nextBound = (nextIndex > stride ?
nextIndex - stride : 0))) {
bound = nextBound;
i = nextIndex - 1;
advance = false;
}
}
if (i < 0 || i >= n || i + n >= nextn) {
int sc;
if (finishing) {
nextTable = null;
table = nextTab;
sizeCtl = (n << 1) - (n >>> 1);
return;
}
if (U.compareAndSwapInt(this, SIZECTL, sc = sizeCtl, sc - 1)) {
if ((sc - 2) != resizeStamp(n) << RESIZE_STAMP_SHIFT)
return;
finishing = advance = true;
i = n; // recheck before commit
}
}
// 如果当前数组i位置元素为空，则创建一个fwd元素，那么则扩容前进
else if ((f = tabAt(tab, i)) == null)
advance = casTabAt(tab, i, null, fwd);
// 如果i位置上的hash等于MOVED表示有其他线程正在扩容，那么则扩容前进
else if ((fh = f.hash) == MOVED)
advance = true; // already processed
else {
// 如果i位置上有元素，那么先锁住这个位置，然后进行扩容
synchronized (f) {
if (tabAt(tab, i) == f) {
Node<K,V> ln, hn;
// 一种是链表的扩容转移
if (fh >= 0) {
int runBit = fh & n;
Node<K,V> lastRun = f;
for (Node<K,V> p = f.next; p != null; p = p.next) {
int b = p.hash & n;
if (b != runBit) {
runBit = b;
lastRun = p;
}
}
if (runBit == 0) {
ln = lastRun;
hn = null;
}
else {
hn = lastRun;
ln = null;
}
for (Node<K,V> p = f; p != lastRun; p = p.next) {
int ph = p.hash; K pk = p.key; V pv = p.val;
if ((ph & n) == 0)
ln = new Node<K,V>(ph, pk, pv, ln);
else
hn = new Node<K,V>(ph, pk, pv, hn);
}
setTabAt(nextTab, i, ln);
setTabAt(nextTab, i + n, hn);
// 转移完之后将i位置元素设置为fwd
setTabAt(tab, i, fwd);
advance = true;
}
// 另外一种是树下面的扩容
else if (f instanceof TreeBin) {
TreeBin<K,V> t = (TreeBin<K,V>)f;
TreeNode<K,V> lo = null, loTail = null;
TreeNode<K,V> hi = null, hiTail = null;
int lc = 0, hc = 0;
for (Node<K,V> e = t.first; e != null; e = e.next) {
int h = e.hash;
TreeNode<K,V> p = new TreeNode<K,V>
(h, e.key, e.val, null, null);
if ((h & n) == 0) {
if ((p.prev = loTail) == null)
lo = p;
else
loTail.next = p;
loTail = p;
++lc;
}
else {
if ((p.prev = hiTail) == null)
hi = p;
else
hiTail.next = p;
hiTail = p;
++hc;
}
}
ln = (lc <= UNTREEIFY_THRESHOLD) ? untreeify(lo) :
(hc != 0) ? new TreeBin<K,V>(lo) : t;
hn = (hc <= UNTREEIFY_THRESHOLD) ? untreeify(hi) :
(lc != 0) ? new TreeBin<K,V>(hi) : t;
setTabAt(nextTab, i, ln);
setTabAt(nextTab, i + n, hn);
// 转移完之后将i位置元素设置为fwd
setTabAt(tab, i, fwd);
advance = true;
}
}
}
}
}
}
7.继续看一下get方法
public V get(Object key) {
Node<K,V>[] tab; Node<K,V> e, p; int n, eh; K ek;
// 计算key的hash值
int h = spread(key.hashCode());
// 根据hash值算出数组对应的下标，然后获取对应位置的值是否为空
if ((tab = table) != null && (n = tab.length) > 0 &&
(e = tabAt(tab, (n - 1) & h)) != null) {
// 如果是当前数组i位置的元素，则直接返回
if ((eh = e.hash) == h) {
if ((ek = e.key) == key || (ek != null && key.equals(ek)))
return e.val;
}
//hash值为负值表示正在扩容，这个时候查的是ForwardingNode的find方法来定位到nextTable来
else if (eh < 0)
return (p = e.find(h, key)) != null ? p.val : null;
// 遍历查找元素
while ((e = e.next) != null) {
if (e.hash == h &&
((ek = e.key) == key || (ek != null && key.equals(ek))))
return e.val;
}
}
return null;
}
8.在继续查看一下remove方法
public V remove(Object key) {
return replaceNode(key, null, null);
}
final V replaceNode(Object key, V value, Object cv) {
int hash = spread(key.hashCode());
// 遍历数组，自旋操作
for (Node<K,V>[] tab = table;;) {
Node<K,V> f; int n, i, fh;
// 如果数组为空，或者数组长度为0，或者key对应位置的元素为空，则跳出循环
if (tab == null || (n = tab.length) == 0 ||
(f = tabAt(tab, i = (n - 1) & hash)) == null)
break;
// 如果key对应位置的元素正在扩容，则帮助扩容
else if ((fh = f.hash) == MOVED)
tab = helpTransfer(tab, f);
else {
V oldVal = null;
boolean validated = false;
// 然后锁住数组上的f元素，进行删除操作
synchronized (f) {
if (tabAt(tab, i) == f) {
// 链表删除逻辑
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;
}
}
// 红黑树的删除逻辑
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));
}
}
}
}
}
// 返回旧值，并进行减1操作
if (validated) {
if (oldVal != null) {
if (value == null)
addCount(-1L, -1);
return oldVal;
}
break;
}
}
}
return null;
}
9.再继续看一下size方法
public int size() {
long n = sumCount();
return ((n < 0L) ? 0 :
(n > (long)Integer.MAX_VALUE) ? Integer.MAX_VALUE :
(int)n);
}
//计算表中的元素总个数
final long sumCount() {
CounterCell[] as = counterCells; CounterCell a;
//baseCount，以这个值作为累加基准
long sum = baseCount;
if (as != null) {
//遍历 counterCells 数组，得到每个对象中的value值
for (int i = 0; i < as.length; ++i) {
if ((a = as[i]) != null)
//累加 value 值
sum += a.value;
}
}
//此时得到的就是元素总个数
return sum;
}
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