ConcurrentHashMap 中putIfAbsent 和put的区别

putIfAbsent 源代码

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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源代码

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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);   }

　　前面一段都是一样的，都是先计算hash再同步取值，区别在于 

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1	s.put(key, hash, value, true); 和

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1	return s.put(key, hash, value, false);<br><br>

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1	final V put(K key, int hash, V value, boolean onlyIfAbsent) {

　　

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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下不会进入修改e.value, 在key已经存在的情况下++modCount; } break; } e = e.next; }

　　

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1	onlyIfAbsent 参数，如果key存在的情况下，在putIfAbsent下不会修改,而put下则会修改成新的值<br>测试put

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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"));

　　结果为：

null
1
2

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1	测试putIfAbsent

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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"));

　结果为:

null
1
1