【源码解读】JDK1.8 中 ConcurrentHashMap 不支持空键值对源码剖析

Original 素文宅博客 Java精选 2022-08-09

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首先明确一点HashMap是支持空键值对的，也就是null键和null值，而ConcurrentHashMap是不支持空键值对的。

查看一下JDK1.8源码，HashMap类部分源码，代码如下：

public V get(Object key) {
        Node<K,V> e;
        return (e = getNode(hash(key), key)) == null ? null : e.value;
}

static final int hash(Object key) {
	int h;
	return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

HashMap在调用put()方法存储数据时会调用hash()方法来计算key的hashcode值，可以从hash()方法上得出当key==null时返回值是0，这意思就是key值是null时，hash()方法返回值是0，不会再调用key.hashcode()方法。

ConcurrentHashMap类部分源码，代码如下：

public V put(K key, V value) {
	return putVal(key, value, false);
}

/** Implementation for put and putIfAbsent */
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();
		else if ((f = tabAt(tab, i = (n - 1) & hash)) == null) {
			if (casTabAt(tab, i, null,
						 new Node<K,V>(hash, key, value, null)))
				break;                   // no lock when adding to empty bin
		}
		else if ((fh = f.hash) == MOVED)
			tab = helpTransfer(tab, f);
		else {
			V oldVal = null;
			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;
			}
		}
	}
	addCount(1L, binCount);
	return null;
}