一.初始化:
hashMap有四种初始化方式:
public HashMap(int initialCapacity, float loadFactor) { if (initialCapacity < 0) throw new IllegalArgumentException("Illegal initial capacity: " + initialCapacity); if (initialCapacity > MAXIMUM_CAPACITY) initialCapacity = MAXIMUM_CAPACITY; if (loadFactor <= 0 || Float.isNaN(loadFactor)) throw new IllegalArgumentException("Illegal load factor: " + loadFactor); this.loadFactor = loadFactor; this.threshold = tableSizeFor(initialCapacity); } public HashMap(int initialCapacity) { this(initialCapacity, DEFAULT_LOAD_FACTOR); } public HashMap() { this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted } public HashMap(Map<? extends K, ? extends V> m) { this.loadFactor = DEFAULT_LOAD_FACTOR; putMapEntries(m, false); }
final void putMapEntries(Map<? extends K, ? extends V> m, boolean evict) { int s = m.size(); if (s > 0) {
// 判断table是否已经初始化
if (table == null) { // pre-size
// 未初始化,s为m的实际元素个数
float ft = ((float)s / loadFactor) + 1.0F; int t = ((ft < (float)MAXIMUM_CAPACITY) ? (int)ft : MAXIMUM_CAPACITY);
// 计算得到的t大于阈值,则初始化阈值, 小疑问: 这里为什么不是threshold = tableSizeFor(t) * loadFactor;?--------------------
if (t > threshold) threshold = tableSizeFor(t); }
// 已初始化,并且m元素个数大于阈值,进行扩容处理
else if (s > threshold) resize();
// 将m中的所有元素添加至HashMap中
for (Map.Entry<? extends K, ? extends V> e : m.entrySet()) { K key = e.getKey(); V value = e.getValue(); putVal(hash(key), key, value, false, evict); } }}
static final int tableSizeFor(int cap) { int n = cap - 1;//为了防止cap已经是2的幂,返回的capacity将是这个cap的两倍的情况出现(例:cap=10000000, (不减1,)经过下面数次无符号右移变成cap+1111111 近似2*cap;).
//MAXIMUM_CAPACITY=1 << 30
//n最大也就只有32bit(一个二进制数据0或者1,是一个bit),这时已经大于MAXIMUM_CAPACITY,所以取MAXIMUM_CAPACITY
n |= n >>> 1; n |= n >>> 2; n |= n >>> 4; n |= n >>> 8; n |= n >>> 16; return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;}
final Node<K,V>[] resize() { Node<K,V>[] oldTab = table;//cap :容量 ; thr: 临界值; int oldCap = (oldTab == null) ? 0 : oldTab.length; int oldThr = threshold; int newCap, newThr = 0; if (oldCap > 0) { if (oldCap >= MAXIMUM_CAPACITY) { threshold = Integer.MAX_VALUE; return oldTab; }// 容量和临界值 都翻倍 else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY && oldCap >= DEFAULT_INITIAL_CAPACITY) newThr = oldThr << 1; // double threshold } else if (oldThr > 0) // initial capacity was placed in threshold newCap = oldThr; else { // zero initial threshold signifies using defaults newCap = DEFAULT_INITIAL_CAPACITY; newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY); }//确保 临界值= 容量*负载因子,并且 临界值 小于MAXIMUM_CAPACITY;
if (newThr == 0) { float ft = (float)newCap * loadFactor; newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ? (int)ft : Integer.MAX_VALUE); }//在此处 初始化 临界值 threshold = newThr; @SuppressWarnings({"rawtypes","unchecked"}) Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap]; table = newTab;//重新散列 hashMap中的元素位置 if (oldTab != null) { for (int j = 0; j < oldCap; ++j) { Node<K,V> e; if ((e = oldTab[j]) != null) { oldTab[j] = null; if (e.next == null)// 重新确定 元素位置 newTab[e.hash & (newCap - 1)] = e; else if (e instanceof TreeNode) ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);// 非链节点, 是红黑树节点 else { // preserve order (保持原有顺序) Node<K,V> loHead = null, loTail = null; Node<K,V> hiHead = null, hiTail = null; Node<K,V> next; do { next = e.next;//
将同一桶中的元素根据(e.hash & oldCap)是否为0进行分割,分成两个不同的链表,完成rehash(若结果为0,则说明扩容后,newCap新增的一个bit=1 对应的 hash值的 位置的值为0【因此,使用的是 oldCap,而不是 oldCap-1】;)
if ((e.hash & oldCap) == 0) { if (loTail == null) loHead = e; else loTail.next = e; loTail = e; } else { if (hiTail == null) hiHead = e; else hiTail.next = e; hiTail = e; } } while ((e = next) != null); if (loTail != null) { loTail.next = null; newTab[j] = loHead; } if (hiTail != null) { hiTail.next = null; newTab[j + oldCap] = hiHead; } } } } } return newTab;}
以下,对扩容方法resize进行补充描述:
通过观测可以发现,我们使用的是2次幂的扩展(长度扩展为原来的2倍),所以,元素的位置要么是原位置,要么是原位置再移动2次幂的位置。其中 n代表 数组长度(容量)。见下图:
因此,元素再重新计算hash之后,因为n变为2倍,那么n-1的二进制表示,在高位会多1bit(如上图),因此新的index就会 = 原位置+orldCap,下为示例:
因此,在扩容时,不需要像旧版本那样重新计算hash,只需要看看原来的hash值新增的那个bit是1还是0就好了,是0的话索引不变,是1的话索引变成原索引+oldCap.(逻辑与运算)
这样设计后,既能省去了重新计算hash值的时间,而且由于新增的1bit是0还是1可以认为是随机的.因此resize的过程,均匀的把之前冲突的节点分散到新的bucket(数组中的位置)了.
且通过比较旧版本,可以发现不同:扩容后,链表元素位置并没有如旧版本一样发生链表中元素倒置的现象,仍然采用的旧的顺序.
这里应该有一个bug,如下:
do { next = e.next; if ((e.hash & oldCap) == 0) { if (loTail == null) loHead = e; else loTail.next = e; //正确的应该是:loHead.next = e; loTail = e; } else { if (hiTail == null) hiHead = e; else hiTail.next = e; //正确的应该是:hiHead.next = e; hiTail = e; } } while ((e = next) != null);
此处作者的本意应该是将旧的链 按 规则(e.hash & oldCap == 0)拆分成两条链, 如此,应该分别使用loHead.next = e 和 hiHead.next = e;而不是loTail.next = e 和hiTail.next = e.
晚些时候,我来写段代码测一测(待续).
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);//红黑树 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 //当链的长度大于等于8,则 这条 链 会被 转换为 红黑树链(其它链不会收到影响) 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; }
今天先到这里,后面继续....
时间: 2024-08-08 09:35:27