大厂面试题：我们知道HashMap是线程不安全的，请编码写一个不安全的案例并给出解决方案？
1、HashMap线程不安全问题产生
import java.util.*;
public class ContainerNotSafeHashMapDemoThree {
public static void main(String[] args) {
Map<String, String> map = new HashMap<>();
for (int i = 0; i < 30; i++) {
new Thread(() -> {
map.put(Thread.currentThread().getName(), UUID.randomUUID().toString());
System.out.println(map.toString());
}, String.valueOf(i)).start();
}
}
}
程序执行结果如下：报java.util.ConcurrentModificationException异常
2、ConcurrentModificationException产生的原因
一个线程正在写，另一个线程过来抢占资源，会造成数据不一致，进而报并发修改异常。
3、HashMap线程不安全解决方案
（1）第一种解决方案
使用Collections工具类创建同步集合类
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
import java.util.UUID;
import java.util.concurrent.ConcurrentHashMap;
public class ContainerSafeHashMapDemoOne {
public static void main(String[] args) {
Map<String, String> map = Collections.synchronizedMap(new HashMap<>());
for (int i = 0; i < 30; i++) {
new Thread(() -> {
map.put(Thread.currentThread().getName(), UUID.randomUUID().toString());
System.out.println(map.toString());
}, String.valueOf(i)).start();
}
}
}
（2）第二种解决方案
使用并发编程类ConcurrentHashMap替换HashMap
import java.util.Map;
import java.util.Set;
import java.util.UUID;
import java.util.concurrent.ConcurrentHashMap;
public class ContainerSafeHashMapDemoTwo {
public static void main(String[] args) {
Map<String, String> map = new ConcurrentHashMap<>();
for (int i = 0; i < 30; i++) {
new Thread(() -> {
map.put(Thread.currentThread().getName(), UUID.randomUUID().toString());
System.out.println(map.toString());
}, String.valueOf(i)).start();
}
}
}
ConcurentHashMap底层使用的是分段锁
public V put(K key, V value) {
return putVal(key, value, false);
}
/** Implementation for put and putIfAbsent */
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; // 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;
}