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HashMap
, HashSet.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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