ConcurrentHashMap 中putIfAbsent 和put的区别

putIfAbsent 源代码

    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

  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 

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

  

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
测试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"));

  结果为:

null
1
2

测试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"));

 结果为:

null
1
1

 

posted @ 2017-03-02 11:28  谭志宇  阅读(17110)  评论(0编辑  收藏  举报