ConcurrentHashMap 中putIfAbsent 和put的区别

Posted 2020-09-02 风雷

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