HashSet集合保证元素唯一性的源码分析(睡起来再看一遍
作者:互联网
//创建集合对象 HashSet<String> hs = new HashSet<>(); //添加元素 hs.add("hello"); hs.add("world"); hs.add("java"); public boolean add(E e) {// 1 return map.put(e, PRESENT)==null; } static final int hash(Object key) {// 3 int h; return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16); } public V put(K key, V value) {// 2 return putVal(hash(key), key, value, false, true); } //hash值和元素的hasCode()方法相关 final V putVal(int hash, K key, V value, boolean onlyIfAbsent,// 4 boolean evict) { Node<K,V>[] tab; Node<K,V> p; int n, i; //如果哈希表未初始化,就对其进行初始化 if ((tab = table) == null || (n = tab.length) == 0) n = (tab = resize()).length; //根据对象的哈希值计算对象的存储位置,如果该位置没有元素,就存储元素 if ((p = tab[i = (n - 1) & hash]) == null) tab[i] = newNode(hash, key, value, null); else { Node<K,V> e; K k; /* ★ 存入的元素和以前的元素比较哈希值 //① 如果哈希值不同,会继续向下执行,把元素添加到集合//② 如果哈希值相同,会调用对象的equals()方法比较//③ 如果返回false,会继续向下执行,把元素添加到集合//④ 如果返回true,说明元素重复,不存储//⑤ */ if (p.hash == hash &&//① ((k = p.key) == key || (key != null && key.equals(k))))//③ e = p;//⑤ 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);//② //④ 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; }
标签:唯一性,hash,tab,HashSet,元素,value,源码,key,null 来源: https://www.cnblogs.com/dayangchongyang/p/15866736.html