1. 集合线程不安全演示
以 ArrayList 为例，我们进入 ArrayList 源码，找到 add() 方法，源码如下
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| public boolean add(E e) {
ensureCapacityInternal(size + 1);
elementData[size++] = e;
return true;
}
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显然，add() 方法没有使用同步互斥，所以在多线程并发时，会出现线程异常，测试代码如下
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| public static void main(String[] args) {
ArrayList<String> list = new ArrayList<>();
for (int i = 0; i < 10; i++) {
new Thread(()->{
list.add(UUID.randomUUID().toString().substring(0,8));
System.out.println(list);
},String.valueOf(i)).start();
}
}
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会出现如下异常
2. 解决方案
2.1 Vector
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| List<String> list = new Vector<>();
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源码如下, 使用 synchronized 修饰，但是这样做效率十分低下且占用资源
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| public synchronized boolean add(E e) {
modCount++;
ensureCapacityHelper(elementCount + 1);
elementData[elementCount++] = e;
return true;
}
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2.2 Collections
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| List<String> list = Collections.synchronizedList(new ArrayList<>());
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进入 Collections 的底层，找到 synchronizedList(List list) 方法，源代码如下，返回指定列表支持的同步（线程安全的）列表
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| public static <T> List<T> synchronizedList(List<T> list) {
return (list instanceof RandomAccess ?
new SynchronizedRandomAccessList<>(list) :
new SynchronizedList<>(list));
}
static <T> List<T> synchronizedList(List<T> list, Object mutex) {
return (list instanceof RandomAccess ?
new SynchronizedRandomAccessList<>(list, mutex) :
new SynchronizedList<>(list, mutex));
}
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2.3 CopyOnWriteArrayList
使用最多，涉及的底层原理为写时复制技术
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| List<String> list = new CopyOnWriteArrayList<>();
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源码如下
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| public boolean add(E e) {
final ReentrantLock lock = this.lock;
lock.lock();
try {
Object[] elements = getArray();
int len = elements.length;
Object[] newElements = Arrays.copyOf(elements, len + 1);
newElements[len] = e;
setArray(newElements);
return true;
} finally {
lock.unlock();
}
}
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对比三者来看，Vector 和 Collections 虽然也可以实现同步，但由于这两种方法在底层都使用了 synchronized 重量级锁，使其效率很低，所以对 ArrayList 的同步主要采用 CopyOnWriteArrayList
3. HashSet的线程不安全
HashSet 同时读写时也会出现 ConcurrentModificationException 异常
其 add() 源代码如下, 没有对做同步处理
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| public boolean add(E e) {
return map.put(e, PRESENT)==null;
}
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其解决方法与 CopyOnWriteArrayList 类似，在 JDK1.8 中，也有一个类叫做 CopyOnWriteArraySet，其底层代码如下
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| public boolean add(E e) {
return al.addIfAbsent(e);
}
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通过 Debug 找到了对关键的一个函数，发现其实现方式与 CopyOnWriteArrayList 底层实现方式类似
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private boolean addIfAbsent(E e, Object[] snapshot) {
final ReentrantLock lock = this.lock;
lock.lock();
try {
Object[] current = getArray();
int len = current.length;
if (snapshot != current) {
int common = Math.min(snapshot.length, len);
for (int i = 0; i < common; i++)
if (current[i] != snapshot[i] && eq(e, current[i]))
return false;
if (indexOf(e, current, common, len) >= 0)
return false;
}
Object[] newElements = Arrays.copyOf(current, len + 1);
newElements[len] = e;
setArray(newElements);
return true;
} finally {
lock.unlock();
}
}
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4. HashMap的线程不安全
HashMap 同时读写时一样会出现 ConcurrentModificationException 异常
进入 HashMap 底层，其 put(K key, V value) 源代码如下
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| 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)
treeifyBin(tab, hash);
break;
}
if (e.hash == hash &&
((k = e.key) == key || (key != null && key.equals(k))))
break;
p = e;
}
}
if (e != null) {
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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在 JDK1.8 中，也有一个类叫做 ConcurrentHashMap ，实现 HashMap 的同步问题，其底层代码如下
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| public V put(K key, V value) {
return putVal(key, value, false);
}
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;
}
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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