Jdk1.8 ConcurrentHashMap 基礎
阿新 • • 發佈：2018-11-22
ConcurrentHashMap
1. ConcurrentHashMap 特性
- key值不能為空,如果為空程式會丟擲空指標異常
- 執行緒安全
- 支援快速失敗(在遍歷的過程中不能進行put或者remove操作)
2. ConcurrentHashMap put過程
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(); // 1.計算傳入key對應的index值,並且判斷是否為空 else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) { //2. 如果為空則通過cas的方式設定值(防止在HashMap中存在的值覆蓋問題),如果設定失敗,則進入下一迴圈 if (casTabAt(tab, i, null, new Node<K,V>(hash, key, value, null))) break; // no lock when adding to empty bin } // 3.如果該node處在擴容的狀態 else if ((fh = f.hash) == MOVED) tab = helpTransfer(tab, f); else { V oldVal = null; // 4.利用synchronized 同步連結串列操作,相比jdk1.7,鎖粒度變小了 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; } } } // 給baseCount +1,並且判斷是否需要擴容 addCount(1L, binCount); return null; }
3. ConcurrentHashMap resize()過程
private final void transfer(Node<K,V>[] tab, Node<K,V>[] nextTab) { int n = tab.length, stride; if ((stride = (NCPU > 1) ? (n >>> 3) / NCPU : n) < MIN_TRANSFER_STRIDE) stride = MIN_TRANSFER_STRIDE; // subdivide range 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); boolean advance = true; boolean finishing = false; // to ensure sweep before committing nextTab for (int i = 0, bound = 0;;) { Node<K,V> f; int fh; 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 } } else if ((f = tabAt(tab, i)) == null) advance = casTabAt(tab, i, null, fwd); else if ((fh = f.hash) == MOVED) advance = true; // already processed else { synchronized (f) { if (tabAt(tab, i) == f) { Node<K,V> ln, hn; 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); setTabAt(tab, i, fwd); advance = true; } 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); setTabAt(tab, i, fwd); advance = true; } } } } } }
4. ConcurrentHashMap 如何統計size
統計size,主要看兩個方法sumCount和addCount.sumCount中遍歷所有的counterCells,然後加上baseCount.為什麼要遍歷所有的counterCells呢?.因為在addCount方法中設定baseCount存在CAS操作失敗的可能性,如果CAS操作失敗則會把count記錄到counterCells中.所有統計size值需要遍歷counterCells的值.
sunCount
final long sumCount() { CounterCell[] as = counterCells; CounterCell a; long sum = baseCount; if (as != null) { for (int i = 0; i < as.length; ++i) { if ((a = as[i]) != null) sum += a.value; } } return sum; }
addCount
private final void addCount(long x, int check) {
CounterCell[] as; long b, s;
if ((as = counterCells) != null ||
!U.compareAndSwapLong(this, BASECOUNT, b = baseCount, s = b + x)) {
CounterCell a; long v; int m;
boolean uncontended = true;
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();
}
if (check >= 0) {
Node<K,V>[] tab, nt; int n, sc;
while (s >= (long)(sc = sizeCtl) && (tab = table) != null &&
(n = tab.length) < MAXIMUM_CAPACITY) {
int rs = resizeStamp(n);
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);
s = sumCount();
}
}
}