The relationship between Hashtable/HashMap and key/value being null

  We all know the three major differences between Hashtable and HashMap, one of which is that HashMap can store a Key is null and multiple elements whose value is null , but Hashtable cannot . Why exactly? Take a look at the source code below:

HashMap.class:

    // 此处计算key的hash值时,会判断是否为null,如果是,则返回0,即key为null的键值对
    // 的hash为0。因此一个hashmap对象只会存储一个key为null的键值对因为它们的hash值都相同,都为0。
    static final int hash(Object key) {
        int h;
        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }
    // 将键值对放入table中时,不会校验value是否为null。因此一个hashmap对象可以存储
    // 多个value为null的键值对
    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }
    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }
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Hashtable.class:

    public synchronized V put(K key, V value) {
        // 确保value不为空。这句代码过滤掉了所有value为null的键值对。因此Hashtable不能
        // 存储value为null的键值对
        if (value == null) {
            throw new NullPointerException();
        }

        // 确保key在table数组中尚未存在。
        Entry<?,?> tab[] = table;
        int hash = key.hashCode(); //在此处计算key的hash值,如果此处key为null,则直接抛出空指针异常。
        int index = (hash & 0x7FFFFFFF) % tab.length;
        @SuppressWarnings("unchecked")
        Entry<K,V> entry = (Entry<K,V>)tab[index];
        for(; entry != null ; entry = entry.next) {
            if ((entry.hash == hash) && entry.key.equals(key)) {
                V old = entry.value;
                entry.value = value;
                return old;
            }
        }

        addEntry(hash, key, value, index);
        return null;
    }
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ConcurrentHashMap.class:

    public V put(K key, V value) {
        return putVal(key, value, false);
    }
    final V putVal(K key, V value, boolean onlyIfAbsent) {
        // 在此处直接过滤掉key或value为null的情况
        if (key == null || value == null) throw new NullPointerException();
        // 另外其hash值采用了二次hash,使得hash值分布更均匀。
        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;
    }
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The analysis of the above code and comments can be summarized as follows:

  1. HashMap calls a separate method when calculating the hash value of the key. In this method, it will judge whether the key is null, and if so, it will return 0; while in Hashtable, the hashCode() method of the key is directly called, so if the key is null, A null pointer exception is thrown.

  2. When HashMap adds key-value pairs into the array, it does not actively determine whether the value is null; while Hashtable first determines whether the value is null.

  3. The above reasons are mainly because Hashtable inherits from Dictionary, and HashMap inherits from AbstractMap.

  4. Although ConcurrentHashMap also inherits from AbstractMap, it also filters out key-value pairs whose key or value is null.


Reprinted from: https://blog.csdn.net/niuwei22007/article/details/52005329

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