zoukankan      html  css  js  c++  java
  • HashMap 源码分析

    本文分析基于 JDK1.8

    HashMap是使用数组 + 链表 + 红黑树 的结构来存储的。当hash冲突时使用链表来存储数据,当链表上的数量大于等于8时,转换成红黑树存储元素。

    hashCode 方法

    HashMap中处处用到了hashCode,先说一下hashCode方法。

    HashCode#hash(Object key)

        /**
         * Computes key.hashCode() and spreads (XORs) higher bits of hash
         * to lower.  Because the table uses power-of-two masking, sets of
         * hashes that vary only in bits above the current mask will
         * always collide. (Among known examples are sets of Float keys
         * holding consecutive whole numbers in small tables.)  So we
         * apply a transform that spreads the impact of higher bits
         * downward. There is a tradeoff between speed, utility, and
         * quality of bit-spreading. Because many common sets of hashes
         * are already reasonably distributed (so don't benefit from
         * spreading), and because we use trees to handle large sets of
         * collisions in bins, we just XOR some shifted bits in the
         * cheapest possible way to reduce systematic lossage, as well as
         * to incorporate impact of the highest bits that would otherwise
         * never be used in index calculations because of table bounds.
         */
        static final int hash(Object key) {
            int h;
            // 用hashCode的高位与低位进行异或运算,让高位参与运算,防止hashCode过小时冲突问题
            return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
        }
    

    知识点补充:

    • ^ 运算是相应位数的什不一样时取1,否则取0
    • 正数在计算机底层是以原码形式存储,而负数是以补码形式存储
    • Integer#hashCode() 值是本身;Long#hashCode()HashMap 的相似 (int)(value ^ (value >>> 32))String#hashCode()h = 31 * h + val[i]; 把原本的 hash 乘以31后再加上每个字符串起来。

    HashMap 的构造器

    无参构造器

        /**
         * The load factor used when none specified in constructor.
         */
        static final float DEFAULT_LOAD_FACTOR = 0.75f;
        
       /**
         * Constructs an empty <tt>HashMap</tt> with the default initial capacity
         * (16) and the default load factor (0.75).
         */
        public HashMap() {
            this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
        }
    

    指定了默认的加载因子,Node 的初始化在第一次放元素的时候。

    有参构造器

    有参构造器最后调用的都是一个。

        /**
         * Constructs an empty <tt>HashMap</tt> with the specified initial
         * capacity and load factor.
         *
         * @param  initialCapacity the initial capacity
         * @param  loadFactor      the load factor
         * @throws IllegalArgumentException if the initial capacity is negative
         *         or the load factor is nonpositive
         */
        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);
        }
    

    注意的是 tableSizeFor(initialCapacity) 的方法,取比自己最大且最近的2的幂次方的数。

        /**
         * Returns a power of two size for the given target capacity.
         */
        static final int tableSizeFor(int cap) {
            // 先减1,防止该数刚好是2的幂次方
            int n = cap - 1;
            // 利用最高位的1把低全变成1,这样变变成全部的1了。比如:9
            // 1001 | 100 = 1101
            n |= n >>> 1;
            // 1101 | 10 = 1111,上次或后最差的情况会把前两位变成1,这一步操作后就能把前4位变成1
            n |= n >>> 2;
            // 1111 | 1 = 1111
            n |= n >>> 4;
            // 1111 | 0 = 1111
            n |= n >>> 8;
            // 1111 | 0 = 1111
            n |= n >>> 16;
            // 因为是以前面或后的结果再移动,所以移动16次刚好把int的数给处理完。
            // 再加1返回,刚好是2的幂次方
            return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
        }
    

    举例说明:

    n = 0100 0000 0000 0000
    
    0100 0000 0000 0000
     010 0000 0000 0000		n |= n >>> 1
    0110 0000 0000 0000
      01 1000 0000 0000		n |= n >>> 2
    0111 1000 0000 0000
         0111 1000 0000		n |= n >>> 4
    

    put方法

    putVal() 方法

        /**
         * 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
         */
    	// onlyIfAbsent key 存在时是否覆盖 value,true为不覆盖
    	// evict HashMap中什么也没有做,在子类LinkHashMap中用来删除头节点。
        final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                       boolean evict) {
            Node<K,V>[] tab; Node<K,V> p; int n, i;
            // 第一次put时,会初始化数组
            if ((tab = table) == null || (n = tab.length) == 0)
                // 初始化 & 扩容 后面分析
                n = (tab = resize()).length;
            if ((p = tab[i = (n - 1) & hash]) == null)
                // 当该位置为null时直接放入该位置
                tab[i] = newNode(hash, key, value, null);
            else {
                // 当该数组的位置不为null时,然后判断是链表还是红黑树
                Node<K,V> e; K k;
                if (p.hash == hash &&
                    ((k = p.key) == key || (key != null && key.equals(k))))
                    // 如果插入元素的key与position上key相等,就把赋值给临时变量e
                    e = p;
                else if (p instanceof TreeNode)
                    // 如果是红黑树节点,就放入树中
                    e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
                else {
                    // key不等于链表的头节点,也不是红黑树时,遍历链表,放到链表的最后面
                    for (int binCount = 0; ; ++binCount) {
                        if ((e = p.next) == null) {
                            // 放置链表最后面
                            p.next = newNode(hash, key, value, null);
                            // >= 7时,即链表上有8个元素的时候进行转换成红黑树
                            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))))
                            // 等于链表上存在的某一个节点时,退出,走下面的是否覆盖的操作
                            break;
                        p = e;
                    }
                }
                if (e != null) { // existing mapping for key
                    // 修改值并返回旧值
                    V oldValue = e.value;
                    if (!onlyIfAbsent || oldValue == null)
                        e.value = value;
                    afterNodeAccess(e);
                    return oldValue;
                }
            }
            ++modCount;
            if (++size > threshold)
                // 扩容
                resize();
            afterNodeInsertion(evict);
            return null;
        }
    

    resize() 方法

        /**
         * Initializes or doubles table size.  If null, allocates in
         * accord with initial capacity target held in field threshold.
         * Otherwise, because we are using power-of-two expansion, the
         * elements from each bin must either stay at same index, or move
         * with a power of two offset in the new table.
         *
         * @return the table
         */
        final Node<K,V>[] resize() {
            Node<K,V>[] oldTab = table;
            int oldCap = (oldTab == null) ? 0 : oldTab.length;
            // 旧数组的阈值
            int oldThr = threshold;
            int newCap, newThr = 0;
            // 扩容而非初始化
            if (oldCap > 0) {
                // 容量达到最大,无法扩容
                if (oldCap >= MAXIMUM_CAPACITY) {
                    threshold = Integer.MAX_VALUE;
                    return oldTab;
                }
                // 扩容
                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;
            // 初始化数组
            else {               // zero initial threshold signifies using defaults
                newCap = DEFAULT_INITIAL_CAPACITY;
                newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
            }
            // 计算扩容的阈值
            if (newThr == 0) {
                float ft = (float)newCap * loadFactor;
                newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
                          (int)ft : Integer.MAX_VALUE);
            }
            threshold = newThr;
            @SuppressWarnings({"rawtypes","unchecked"})
                Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
            table = newTab;
            if (oldTab != null) {
                for (int j = 0; j < oldCap; ++j) {
                    Node<K,V> e;
                    if ((e = oldTab[j]) != null) {
                        oldTab[j] = null;
                        // 该位置上只有一个元素,直接放到新数组
                        if (e.next == null)
                            newTab[e.hash & (newCap - 1)] = e;
                        // 该位置是红黑树结构
                        else if (e instanceof TreeNode)
                            ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                        // 链表
                        else { // preserve order
                            Node<K,V> loHead = null, loTail = null;
                            Node<K,V> hiHead = null, hiTail = null;
                            Node<K,V> next;
                            do {
                                next = e.next;
                                // 在新数组的下标还是当前的下标
                                if ((e.hash & oldCap) == 0) {
                                    if (loTail == null)
                                        loHead = e;
                                    else
                                        loTail.next = e;
                                    loTail = e;
                                }
                                // 在新数组的下标是 oldCap + index
                                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;
        }
    

    链表迁移

    假如旧链表在旧数组的下标是3,capacity 是16,扩容后,capacity 是32。链表迁移是把链表上的数据分成两组,一组数据在新数组的下标是3,另一 组的下标应该是19(oldCapacity + index)。分组的依据是使用 e.hash & oldCap 因为在旧的数组上计算下脚标使用的是 e.hash & oldCap - 1 ,所以扩容后,只比较 oldCap 的最高位就能确定在扩容后数组的位置。

    get方法

    get的代码比较简单。

    get(Object)

        /**
         * Returns the value to which the specified key is mapped,
         * or {@code null} if this map contains no mapping for the key.
         *
         * <p>More formally, if this map contains a mapping from a key
         * {@code k} to a value {@code v} such that {@code (key==null ? k==null :
         * key.equals(k))}, then this method returns {@code v}; otherwise
         * it returns {@code null}.  (There can be at most one such mapping.)
         *
         * <p>A return value of {@code null} does not <i>necessarily</i>
         * indicate that the map contains no mapping for the key; it's also
         * possible that the map explicitly maps the key to {@code null}.
         * The {@link #containsKey containsKey} operation may be used to
         * distinguish these two cases.
         *
         * @see #put(Object, Object)
         */
        public V get(Object key) {
            Node<K,V> e;
            return (e = getNode(hash(key), key)) == null ? null : e.value;
        }
    

    getNode(int, Object)

        /**
         * Implements Map.get and related methods
         *
         * @param hash hash for key
         * @param key the key
         * @return the node, or null if none
         */
        final Node<K,V> getNode(int hash, Object key) {
            Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
            if ((tab = table) != null && (n = tab.length) > 0 &&
                (first = tab[(n - 1) & hash]) != null) {
                // 如果是数组元素(链表/红黑树的第一个值),直接返回
                if (first.hash == hash && // always check first node
                    ((k = first.key) == key || (key != null && key.equals(k))))
                    return first;
                if ((e = first.next) != null) {
                    if (first instanceof TreeNode)
                        // 查找红黑树
                        return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                    do {
                        // 遍历查看链表
                        if (e.hash == hash &&
                            ((k = e.key) == key || (key != null && key.equals(k))))
                            return e;
                    } while ((e = e.next) != null);
                }
            }
            return null;
        }
    
  • 相关阅读:
    华硕路由器修改 Hosts 以达到局域网内自定义解析
    一款开源、高颜值的终端terminus,支持Windows、MacOS
    Windows 10启用Linux子系统(WSL)
    一款全能的下载工具Motrix,支持BT、磁力链、百度网盘等资源
    ubuntu 14.04 和16.04 快速下载
    CentOS 7一键安装Seafile搭建私有云存储
    background背景色
    3d爱心代码
    Mac Mini(late 2014) 添加NVMe固态组Fusion Drive
    member access within misaligned address 0x0000002c3931 for type 'struct ListNode‘
  • 原文地址:https://www.cnblogs.com/chaos-x/p/13377384.html
Copyright © 2011-2022 走看看