5.2.1）HashSet集合线程不安全案例展示
代码如下：
// HashSet集合线程不安全案例展示
public class HashSetErrorDemo {
public static void main(String[] args) {
//演示Hashset
Set set = new HashSet<>();
for (int i = 0; i < 30; i++) {
new Thread(() -> {
//向集合添加内容
set.add(UUID.randomUUID().toString().substring(0, 8));
//从集合获取内容
System.out.println(set);
}, String.valueOf(i)).start();
}
}
}
输出：
java.util.ConcurrentModificationException at java.util.HashMapKeyIterator.next(HashMap.java:1469) at java.util.AbstractCollection.toString(AbstractCollection.java:461) at java.lang.String.valueOf(String.java:2994) at java.io.PrintStream.println(PrintStream.java:821) at com.study.collection.HashSetErrorDemo.lambda0(HashSetErrorDemo.java:25) at java.lang.Thread.run(Thread.java:748) java.util.ConcurrentModificationException at java.util.HashMapKeyIterator.next(HashMap.java:1469) at java.util.AbstractCollection.toString(AbstractCollection.java:461) at java.lang.String.valueOf(String.java:2994) at java.io.PrintStream.println(PrintStream.java:821) at com.study.collection.HashSetErrorDemo.lambda0(HashSetErrorDemo.java:25) at java.lang.Thread.run(Thread.java:748)
发现报错，java.util.ConcurrentModificationException ——》并发修改异常
问题: 为什么会出现并发修改异常?
查看HashSet的add方法源码 ，如下：
public boolean add(E e) {
return map.put(e, PRESENT)==null;
}
add() 方法没有添加 synchronized 关键字，即没有对多线程情况进行处理
5.2.2）HashSet集合线程不安全解决方案
5.2.2.1）解决方案-CopyOnWriteArraySet
使用 CopyOnWriteArraySet解决HashSet集合线程不安全问题，代码如下：
// CopyOnWriteArray解决HashSet集合线程不安全问题
public class CopyOnWriteArraySetDemo {
public static void main(String[] args) {
//演示Hashset
Set set = new CopyOnWriteArraySet<>();
for (int i = 0; i < 30; i++) {
new Thread(() -> {
//向集合添加内容
set.add(UUID.randomUUID().toString().substring(0, 8));
//从集合获取内容
System.out.println(set);
}, String.valueOf(i)).start();
}
}
}
输出：
com.study.collection.HashSet.CopyOnWriteArraySetDemo
[6cfeb375, cc511c07, 79eddeb2, 6ba3f255, edf8d95d, 32cb0924, 5a3189bf, 196e5a54, 9875e04b, e240ad35, 22debfb2, 435bae49, 968d06de, d7fdacae, a46e7686, 6ea6c822, b380ce03, 54ddfa83, 16d0a020] [6cfeb375, cc511c07, 79eddeb2, 6ba3f255, edf8d95d, 32cb0924, 5a3189bf, 196e5a54, 9875e04b, e240ad35, 22debfb2, 435bae49, 968d06de, d7fdacae, a46e7686, 6ea6c822, b380ce03, 54ddfa83, 16d0a020, 2f4befd6]
...
[6cfeb375, cc511c07, 79eddeb2, 6ba3f255, edf8d95d, 32cb0924, 5a3189bf, 196e5a54, 9875e04b, e240ad35, 22debfb2, 435bae49, 968d06de, d7fdacae, a46e7686, 6ea6c822, b380ce03, 54ddfa83, 16d0a020, 2f4befd6, 8d3201e7, b4d2fb7a]
Process finished with exit code 0
CopyOnWriteArraySet的add() 方法源码如下：addIfAbsent()中实现了lock
final transient ReentrantLock lock = new ReentrantLock();
public boolean add(E e) {
return al.addIfAbsent(e);
}
public boolean addIfAbsent(E e) {
Object[] snapshot = getArray();
return indexOf(e, snapshot, 0, snapshot.length) >= 0 ? false :
addIfAbsent(e, snapshot);
}
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) {
// Optimize for lost race to another addXXX operation
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();
}
}
5.3）HashMap集合线程
5.3.1）HashMap集合线程不安全案例展示
代码如下：
// HashMap集合线程不安全案例展示
public class HashMapErrorDemo {
public static void main(String[] args) {
//演示HashMap
Map map = new HashMap<>();
for (int i = 0; i <30; i++) {
String key = String.valueOf(i);
new Thread(()->{
//向集合添加内容
map.put(key,UUID.randomUUID().toString().substring(0,8));
//从集合获取内容
System.out.println(map);
},String.valueOf(i)).start();
}
}
}
输出：
Exception in thread "17" Exception in thread "1" Exception in thread "16" Exception in thread "26" java.util.ConcurrentModificationException at java.util.HashMapEntryIterator.next(HashMap.java:1479) at java.util.HashMapmainHashIterator.nextNode(HashMap.java:1445) at java.util.HashMapEntryIterator.next(HashMap.java:1477) at java.util.AbstractMap.toString(AbstractMap.java:554) at java.lang.String.valueOf(String.java:2994) at java.io.PrintStream.println(PrintStream.java:821) at com.study.collection.HashMap.HashMapErrorDemo.lambda0(HashMapErrorDemo.java:27) at java.lang.Thread.run(Thread.java:748)
发现报错，java.util.ConcurrentModificationException ——》并发修改异常
问题: 为什么会出现并发修改异常?
查看HashMap的put方法源码 ，如下：
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[] tab; Node 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 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)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;
}
put() 方法没有添加 synchronized 关键字，即没有对多线程情况进行处理
5.3.2）HashMap集合线程不安全解决方案
5.3.2.1）解决方案-ConcurrentHashMap
使用 ConcurrentHashMap 解决HashMap集合线程不安全问题，代码如下：
// ConcurrentHashMap解决HashMap集合线程不安全问题
public class ConcurrentHashMapDemo {
public static void main(String[] args) {
//演示ConcurrentHashMap
Map map = new ConcurrentHashMap<>();
for (int i = 0; i <30; i++) {
String key = String.valueOf(i);
new Thread(()->{
//向集合添加内容
map.put(key, UUID.randomUUID().toString().substring(0,8));
//从集合获取内容
System.out.println(map);
},String.valueOf(i)).start();
}
}
}
输出：
{0=8d7844d0, 1=19cc45b9, 2=a6ece204, 3=3878495c, 4=5f6e1594, 5=a88e727a, 6=9cd46cbd, 7=2f507f9a, 8=3e0da179, 9=61c9b227} {11=feff97c8, 0=8d7844d0, 1=19cc45b9, 2=a6ece204, 3=3878495c, 4=5f6e1594, 5=a88e727a, 6=9cd46cbd, 7=2f507f9a, 8=3e0da179, 9=61c9b227, 10=6b7f272f}
...
{0=8d7844d0, 1=19cc45b9, 2=a6ece204, 3=3878495c, 4=5f6e1594, 5=a88e727a, 6=9cd46cbd, 7=2f507f9a, 8=3e0da179, 9=61c9b227, 10=6b7f272f}
ConcurrentHashMap的put() 方法源码如下：putVal()中添加了synchronize
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[] tab = table;;) {
Node 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(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 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 pred = e;
if ((e = e.next) == null) {
pred.next = new Node(hash, key,
value, null);
break;
}
}
}
else if (f instanceof TreeBin) {
Node p;
binCount = 2;
if ((p = ((TreeBin)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;
}
