【源码探索系列】 modCount知多少？

很多面试官都会问modCount是啥？看了本文，我相信你可以解脱了

1.modCount是什么？

相信很多同学都会在List或hashMap近亲系列源码中都会看到这个modCount变量，简言之，从字面意思理解modCount，修改的次数。

2.modCount的作用

通常地，在集合源码中存在这个modCount变量时，基本上可以说明这个类是线程不安全的。这个变量在集合初始化的时候，就将modCount赋值给exceptCound；在集合迭代器中，对modCount与exceptCount进行比较，如果出现两者不相等，说明有其他线程对该集合进行了修改操作（add或remove）,这个时候系统就会抛出CurrentModifactionException异常，这从逻辑上防止了多线程并发操作集合时，迭代器中数据紊乱的局面。

3.HashMap源码角度分析modCount

3.1 map进行put的时候，modCount++

 /**
     * Associates the specified value with the specified key in this map.
     * If the map previously contained a mapping for the key, the old
     * value is replaced.
     *
     * @param key key with which the specified value is to be associated
     * @param value value to be associated with the specified key
     * @return the previous value associated with key, or
     *         null if there was no mapping for key.
     *         (A null return can also indicate that the map
     *         previously associated null with key.)
     */
    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

    /**
     * Implements Map.put and related methods.
     *
     * @param hash hash for key
     * @param key the key
     * @param value the value to put
     * @param onlyIfAbsent if true, don't change existing value
     * @param evict if false, the table is in creation mode.
     * @return previous value, or null if none
     */
    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;
    }

3.2 map删除key的时候，modCount++

/**
     * Removes the mapping for the specified key from this map if present.
     *
     * @param  key key whose mapping is to be removed from the map
     * @return the previous value associated with key, or
     *         null if there was no mapping for key.
     *         (A null return can also indicate that the map
     *         previously associated null with key.)
     */
    public V remove(Object key) {
        Node<K,V> e;
        return (e = removeNode(hash(key), key, null, false, true)) == null ?
            null : e.value;
    }

    /**
     * Implements Map.remove and related methods.
     *
     * @param hash hash for key
     * @param key the key
     * @param value the value to match if matchValue, else ignored
     * @param matchValue if true only remove if value is equal
     * @param movable if false do not move other nodes while removing
     * @return the node, or null if none
     */
    final Node<K,V> removeNode(int hash, Object key, Object value,
                               boolean matchValue, boolean movable) {
        Node<K,V>[] tab; Node<K,V> p; int n, index;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (p = tab[index = (n - 1) & hash]) != null) {
            Node<K,V> node = null, e; K k; V v;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                node = p;
            else if ((e = p.next) != null) {
                if (p instanceof TreeNode)
                    node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
                else {
                    do {
                        if (e.hash == hash &&
                            ((k = e.key) == key ||
                             (key != null && key.equals(k)))) {
                            node = e;
                            break;
                        }
                        p = e;
                    } while ((e = e.next) != null);
                }
            }
            if (node != null && (!matchValue || (v = node.value) == value ||
                                 (value != null && value.equals(v)))) {
                if (node instanceof TreeNode)
                    ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
                else if (node == p)
                    tab[index] = node.next;
                else
                    p.next = node.next;
                ++modCount;
                --size;
                afterNodeRemoval(node);
                return node;
            }
        }
        return null;
    }

3.3 map进行迭代，modCount与expertCount比较

/**
     * Returns a {@link Set} view of the keys contained in this map.
     * The set is backed by the map, so changes to the map are
     * reflected in the set, and vice-versa.  If the map is modified
     * while an iteration over the set is in progress (except through
     * the iterator's own remove operation), the results of
     * the iteration are undefined.  The set supports element removal,
     * which removes the corresponding mapping from the map, via the
     * Iterator.remove, Set.remove,
     * removeAll, retainAll, and clear
     * operations.  It does not support the add or addAll
     * operations.
     *
     * @return a set view of the keys contained in this map
     */
    public Set<K> keySet() {
        Set<K> ks = keySet;
        if (ks == null) {
            ks = new KeySet();
            keySet = ks;
        }
        return ks;
    }

    final class KeySet extends AbstractSet<K> {
        public final int size()                 { return size; }
        public final void clear()               { HashMap.this.clear(); }
        public final Iterator<K> iterator()     { return new KeyIterator(); }
        public final boolean contains(Object o) { return containsKey(o); }
        public final boolean remove(Object key) {
            return removeNode(hash(key), key, null, false, true) != null;
        }
        public final Spliterator<K> spliterator() {
            return new KeySpliterator<>(HashMap.this, 0, -1, 0, 0);
        }
        public final void forEach(Consumer<? super K> action) {
            Node<K,V>[] tab;
            if (action == null)
                throw new NullPointerException();
            if (size > 0 && (tab = table) != null) {
                int mc = modCount;
                for (int i = 0; i < tab.length; ++i) {
                    for (Node<K,V> e = tab[i]; e != null; e = e.next)
                        action.accept(e.key);
                }
                if (modCount != mc)
                    throw new ConcurrentModificationException();
            }
        }
    }

在迭代之前，先获取expertCount，然后遍历完了之后expertCount与modCount进行比较，不一致就抛
ConcurrentModificationException。

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