背景
在和朋友的一次交谈中，朋友遇到了这样一个问题：在遍历map时，做了一步remove操作，然后就发生了ConcurrentModificationException异常。由此我点进源码探究了一番。
JDK版本：1.8
探究
点进HashMap源码发现，在每次循环结束前，会校验一下modCount的值，如果modCount变了，就会抛出ConcurrentModificationException异常。
@Override
public void forEach(BiConsumer<? super K, ? super V> 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, e.value);
}
if (modCount != mc)
throw new ConcurrentModificationException();
}
}
那么modCount又是什么呢，结合网上搜索出来的以及源码注释的描述，这个参数是用来在HashMap迭代时，可以快速发现错误并结束迭代的。
/**
* The number of times this HashMap has been structurally modified
* Structural modifications are those that change the number of mappings in
* the HashMap or otherwise modify its internal structure (e.g.,
* rehash). This field is used to make iterators on Collection-views of
* the HashMap fail-fast. (See ConcurrentModificationException).
*/
transient int modCount;
知道这个参数的含义后，点击modCount，查看有哪些地方会修改他的值。发现有这些地方会修改modCount。最后赋值为0的方法（reinitialize）是重新初始化HashMap的参数，这个可以略过。
putVal
661行，也是put()实际调用的方法。从源码中我们可以看到，当key存在时，会直接return oldValue结束方法。也就是说只有在插入新的key时会导致modCount增加。
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;
}
removeNode
845行，也是remove()实际调用的方法。在有数据被删除时会修改modCount。
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;
}
clear
860行。把map里面每一个node都会重置为null。并在调用时就会修改modCount的值。
public void clear() {
Node<K,V>[] tab;
modCount++;
if ((tab = table) != null && size > 0) {
size = 0;
for (int i = 0; i < tab.length; ++i)
tab[i] = null;
}
}
computeIfAbsent
1142行。该方法会检测key是否存在，如果不存在则添加且modCount++。
@Override
public V computeIfAbsent(K key, Function<? super K, ? extends V> mappingFunction) {
if (mappingFunction == null)
throw new NullPointerException();
int hash = hash(key);
Node<K,V>[] tab; Node<K,V> first; int n, i;
int binCount = 0;
TreeNode<K,V> t = null;
Node<K,V> old = null;
if (size > threshold || (tab = table) == null ||
(n = tab.length) == 0)
n = (tab = resize()).length;
if ((first = tab[i = (n - 1) & hash]) != null) {
if (first instanceof TreeNode)
old = (t = (TreeNode<K,V>)first).getTreeNode(hash, key);
else {
Node<K,V> e = first; K k;
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k)))) {
old = e;
break;
}
++binCount;
} while ((e = e.next) != null);
}
V oldValue;
if (old != null && (oldValue = old.value) != null) {
afterNodeAccess(old);
return oldValue;
}
}
V v = mappingFunction.apply(key);
if (v == null) {
return null;
} else if (old != null) {
old.value = v;
afterNodeAccess(old);
return v;
}
else if (t != null)
t.putTreeVal(this, tab, hash, key, v);
else {
tab[i] = newNode(hash, key, v, first);
if (binCount >= TREEIFY_THRESHOLD - 1)
treeifyBin(tab, hash);
}
++modCount;
++size;
afterNodeInsertion(true);
return v;
}
compute
1214行。与computeIfAbsen()类似，只会在key不存在且remappingFunction返回的value不为空时进行添加并modCount++。
@Override
public V compute(K key,
BiFunction<? super K, ? super V, ? extends V> remappingFunction) {
if (remappingFunction == null)
throw new NullPointerException();
int hash = hash(key);
Node<K,V>[] tab; Node<K,V> first; int n, i;
int binCount = 0;
TreeNode<K,V> t = null;
Node<K,V> old = null;
if (size > threshold || (tab = table) == null ||
(n = tab.length) == 0)
n = (tab = resize()).length;
if ((first = tab[i = (n - 1) & hash]) != null) {
if (first instanceof TreeNode)
old = (t = (TreeNode<K,V>)first).getTreeNode(hash, key);
else {
Node<K,V> e = first; K k;
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k)))) {
old = e;
break;
}
++binCount;
} while ((e = e.next) != null);
}
}
V oldValue = (old == null) ? null : old.value;
V v = remappingFunction.apply(key, oldValue);
if (old != null) {
if (v != null) {
old.value = v;
afterNodeAccess(old);
}
else
removeNode(hash, key, null, false, true);
}
else if (v != null) {
if (t != null)
t.putTreeVal(this, tab, hash, key, v);
else {
tab[i] = newNode(hash, key, v, first);
if (binCount >= TREEIFY_THRESHOLD - 1)
treeifyBin(tab, hash);
}
++modCount;
++size;
afterNodeInsertion(true);
}
return v;
}
merge
1273行。如果key不存在，则会添加key-value且modCount++。
@Override
public V merge(K key, V value,
BiFunction<? super V, ? super V, ? extends V> remappingFunction) {
if (value == null)
throw new NullPointerException();
if (remappingFunction == null)
throw new NullPointerException();
int hash = hash(key);
Node<K,V>[] tab; Node<K,V> first; int n, i;
int binCount = 0;
TreeNode<K,V> t = null;
Node<K,V> old = null;
if (size > threshold || (tab = table) == null ||
(n = tab.length) == 0)
n = (tab = resize()).length;
if ((first = tab[i = (n - 1) & hash]) != null) {
if (first instanceof TreeNode)
old = (t = (TreeNode<K,V>)first).getTreeNode(hash, key);
else {
Node<K,V> e = first; K k;
do {
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k)))) {
old = e;
break;
}
++binCount;
} while ((e = e.next) != null);
}
}
if (old != null) {
V v;
if (old.value != null)
v = remappingFunction.apply(old.value, value);
else
v = value;
if (v != null) {
old.value = v;
afterNodeAccess(old);
}
else
removeNode(hash, key, null, false, true);
return v;
}
if (value != null) {
if (t != null)
t.putTreeVal(this, tab, hash, key, value);
else {
tab[i] = newNode(hash, key, value, first);
if (binCount >= TREEIFY_THRESHOLD - 1)
treeifyBin(tab, hash);
}
++modCount;
++size;
afterNodeInsertion(true);
}
return value;
}
总结
HashMap当中有7处地方对modCount进行了重新赋值，其中reinitialize和clear方法相当于重置了整个HashMap，removeNode是删除了数据，其他的都是新增数据。在有这些操作的时候，会导致modCount变化。如果在遍历HashMap时，进行过如上操作会导致ConcurrentModificationException异常。
