HashSet.add覆盖问题

Posted by Codeboy on July 4, 2019

背景

假定对象A、B的hash值相同,equals方法也想等,那么向 HashSet 中顺序添加A、B,最后集合中保留的是A或者B或者是A和B呢?

编码

看以下代码,分析下输出:

  • package me.codeboy.test.hash;
  • import java.util.Objects;
  • /**
  • * hash bean
  • * Created by yuedong.li on 2019-07-01
  • */
  • public class HashBean {
  • String value;
  • @Override
  • public boolean equals(Object obj) {
  • return true;
  • }
  • @Override
  • public int hashCode() {
  • return Objects.hash("test");
  • }
  • @Override
  • public String toString() {
  • return value;
  • }
  • }
  • package me.codeboy.test.hash;
  • import com.google.common.collect.Maps;
  • import java.util.HashSet;
  • import java.util.Map;
  • /**
  • * set add问题
  • * Created by yuedong.li on 2019-07-01
  • */
  • public class SetAdd {
  • public static void main(String[] args) {
  • HashBean bean = new HashBean();
  • bean.value = "first";
  • HashBean bean2 = new HashBean();
  • bean2.value = "second";
  • HashSet<HashBean> set = new HashSet<>();
  • set.add(bean);
  • set.add(bean2);
  • System.out.println(set.size());
  • System.out.println(set.iterator().next().value);
  • System.out.println();
  • Map<HashBean, String> map2 = Maps.newHashMap();
  • map2.put(bean, "first");
  • map2.put(bean2, "second");
  • System.out.println(map2.size());
  • System.out.println(map2.values().iterator().next());
  • System.out.println(map2.keySet().iterator().next());
  • }
  • }

分析

这里先贴一下输出的结果:

1
first

1
second
first

为什么会是这样的呢?我们先看一下Jdk1.8.0中 HashSet.add 方法的调用栈:

  • ## HashSet
  • public boolean add(E e) {
  • return map.put(e, PRESENT)==null; //这里直接使用的是hashMap,将值当作key记录
  • }
  • ## HashMap
  • /**
  • * 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 <tt>key</tt>, or
  • * <tt>null</tt> if there was no mapping for <tt>key</tt>.
  • * (A <tt>null</tt> return can also indicate that the map
  • * previously associated <tt>null</tt> with <tt>key</tt>.)
  • */
  • public V put(K key, V value) {
  • return putVal(hash(key), key, value, false, true);
  • }
  • /**
  • * 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的);
  • return null;
  • }

从代码中可以看出,HashSet.add 直接调用HashMap.put 方法,HashSet的内部实现也确实是HashMapHashSet.add 的值直接作为HashMap的key进行存储,从HashMap.putVal 方法中可以看出,HashMap的key并没有做替换,在 onlyIfAbsent是false或者原先值为null的情况下,新value会替换旧value。

小结

HashSet在add的时候,在两个对象相等的情况下,是不进行替换的。

HashMap在put的时候,在两个key相等的情况下,是不进行替换的,在两个value相同的情况下,是要根据 putVal 方法中的onlyIfAbsent字段进行决定的。

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HashSet.add覆盖问题