HashSet是一个常见的集合类，底层使用HashMap实现。那么HashSet如何保证元素不重复呢？如果多次放入equals()为true的相同元素会覆盖吗？今天来探究一下源码
首先，来看看HashSet的无参构造函数
/**
* Constructs a new, empty set; the backing {@code HashMap} instance has
* default initial capacity (16) and load factor (0.75).
*/
public HashSet() {
map = new HashMap<>();
}
很简单，new一个HashMap。
当然了，HashSet也提供了其他构造函数，指定初始容量和装载因子等，也是调用了HashMap的相应构造方法。
/**
* Constructs a new, empty set; the backing {@code HashMap} instance has
* the specified initial capacity and the specified load factor.
*
* @param initialCapacity the initial capacity of the hash map
* @param loadFactor the load factor of the hash map
* @throws IllegalArgumentException if the initial capacity is less
* than zero, or if the load factor is nonpositive
*/
public HashSet(int initialCapacity, float loadFactor) {
map = new HashMap<>(initialCapacity, loadFactor);
}
/**
* Constructs a new, empty set; the backing {@code HashMap} instance has
* the specified initial capacity and default load factor (0.75).
*
* @param initialCapacity the initial capacity of the hash table
* @throws IllegalArgumentException if the initial capacity is less
* than zero
*/
public HashSet(int initialCapacity) {
map = new HashMap<>(initialCapacity);
}
/**
* Constructs a new, empty linked hash set. (This package private
* constructor is only used by LinkedHashSet.) The backing
* HashMap instance is a LinkedHashMap with the specified initial
* capacity and the specified load factor.
*
* @param initialCapacity the initial capacity of the hash map
* @param loadFactor the load factor of the hash map
* @param dummy ignored (distinguishes this
* constructor from other int, float constructor.)
* @throws IllegalArgumentException if the initial capacity is less
* than zero, or if the load factor is nonpositive
*/
HashSet(int initialCapacity, float loadFactor, boolean dummy) {
map = new LinkedHashMap<>(initialCapacity, loadFactor);
}
再来看看HashSet如何添加元素？
/**
* Adds the specified element to this set if it is not already present.
* More formally, adds the specified element {@code e} to this set if
* this set contains no element {@code e2} such that
* {@code Objects.equals(e, e2)}.
* If this set already contains the element, the call leaves the set
* unchanged and returns {@code false}.
*
* @param e element to be added to this set
* @return {@code true} if this set did not already contain the specified
* element
*/
public boolean add(E e) {
return map.put(e, PRESENT)==null;
}
向map中添加元素，元素作为key，PRESENT作为value，返回是否添加成功。map中key唯一，因此set中的元素也唯一了。这里不免有疑问，PRESENT是什么呢？来看看PRESENT的定义。
// Dummy value to associate with an Object in the backing Map
private static final Object PRESENT = new Object();
原来PRESENT是一个常量，一个空的Object类型实例对象，只是为了占住value的位置，并无实际作用。
接下来看看HashMap的put方法。
/**
* 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 {@code key}, or
* {@code null} if there was no mapping for {@code key}.
* (A {@code null} return can also indicate that the map
* previously associated {@code null} with {@code key}.)
*/
public V put(K key, V value) {
return putVal(hash(key), key, value, false, true);
}
只是简单调用了putVal()方法，继续看putVal()的实现。
/**
* 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;
}
从源码看到，HashMap保证key的不重复性，对于重复的key，HashMap会根据参数onlyIfAbsent的设置和原value是否为空两个条件来判断是否替换新value，但要注意的是，我们前面已经看到，对于HashSet，这个value只是个空的Object类的对象，没有任何实际作用，HashSet中的元素实际上是存储在key上的。针对重复的key，HashMap只有对于value的处理，并不会替换key，因此在HashSet中加入相同元素不会覆盖。
为了更清晰认识这点，我又做了个简单实验，来看看代码。
import java.util.HashSet;
import java.util.Iterator;
import java.util.Set;
public class HashSetTest{
public static void main(String[] args) {
User user1 = new User(1,"abc",20);
User user2= new User(1,"def",20);
Set<User> set = new HashSet<>();
set.add(user1);
set.add(user2);
Iterator<User> iterator = set.iterator();
while(iterator.hasNext()) System.out.println(iterator.next());
}
}
class User{
private int id;
private String name;
private int age;
@Override
public int hashCode() {
return 17*id + 17*age;
}
@Override
public String toString() {
return "User{" +
"id=" + id +
", name='" + name + '\'' +
", age=" + age +
'}';
}
public User(int id, String name, int age) {
this.id = id;
this.name = name;
this.age = age;
}
@Override
public boolean equals(Object obj) {
User userObj = (User) obj;
return this.id==userObj.id && this.age==userObj.age;
}
}
这里，User类有三个成员变量：id、name、age，重写的hashcode()和equals()都只依赖于两个成员变量：id、age，因此在HashSet中，会判断user1、user2是相同对象，我们就能通过name的不同判断是否进行了覆盖。
来看看输出结果：
User{id=1, name='abc', age=20}
到这里，也验证了之前在源码中看到的，向HashSet中加入相同元素不会进行覆盖。因为HashSet底层使用HashMap实现，元素存在HashMap的key中。在HashMap中，多次put相同的key，只会覆盖value，而不存在key的情况。
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