putIfAbsent 源代碼
public V putIfAbsent(K key, V value) {
Segment<K,V> s;
if (value == null)
throw new NullPointerException();
int hash = hash(key);
int j = (hash >>> segmentShift) & segmentMask;
if ((s = (Segment<K,V>)UNSAFE.getObject
(segments, (j << SSHIFT) + SBASE)) == null)
s = ensureSegment(j);
return s.put(key, hash, value, true);
}
put源代碼
public V put(K key, V value) {
Segment<K,V> s;
if (value == null)
throw new NullPointerException();
int hash = hash(key);
int j = (hash >>> segmentShift) & segmentMask;
if ((s = (Segment<K,V>)UNSAFE.getObject // nonvolatile; recheck
(segments, (j << SSHIFT) + SBASE)) == null) // in ensureSegment
s = ensureSegment(j);
return s.put(key, hash, value, false);
}
前面一段都是一樣的,都是先計算hash再同步取值,區別在於
s.put(key, hash, value, true); 和
return s.put(key, hash, value, false);
final V put(K key, int hash, V value, boolean onlyIfAbsent) {
for (HashEntry<K,V> e = first;;) {
if (e != null) {
K k;
if ((k = e.key) == key ||
(e.hash == hash && key.equals(k))) {
oldValue = e.value;
if (!onlyIfAbsent) {
e.value = value; //putIfAbsent下不會進入修改e.value, 在key已經存在的情況下
++modCount; } break; } e = e.next; }
onlyIfAbsent 參數,如果key存在的情況下,在putIfAbsent下不會修改,而put下則會修改成新的值
測試put
ConcurrentHashMap<String, String> map = new ConcurrentHashMap<String, String>();
System.out.println(map.put("1", "1"));
System.out.println(map.put("1", "2"));
System.out.println(map.get("1"));
結果為:
null
1
2
測試putIfAbsent
ConcurrentHashMap<String, String> map = new ConcurrentHashMap<String, String>();
System.out.println(map.putIfAbsent("1", "1"));
System.out.println(map.putIfAbsent("1", "2"));
System.out.println(map.get("1"));
結果為:
null
1
1
