• Java集合--HashMap源码


    本文基于JKD1.8

    1.关键的属性:

        //默认容量
        static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16
        //最大容量
        static final int MAXIMUM_CAPACITY = 1 << 30;
        //默认负载因子
        static final float DEFAULT_LOAD_FACTOR = 0.75f;
        //链表树化的阀值
        static final int TREEIFY_THRESHOLD = 8;
        //树形转为链表的阀值
        static final int UNTREEIFY_THRESHOLD = 6;
        //树化要求的node容量
        static final int MIN_TREEIFY_CAPACITY = 64;
        //实际存放节点的数组
        transient Node<K,V>[] table;
        //KV的Set,相当于一个快照
        transient Set<Map.Entry<K,V>> entrySet;
        //存储的节点的个数
        transient int size;
        //fial-fast机制的modify count个数
        transient int modCount;
        //容量×负载因子得到的阈值,超过该阈值进行扩容
        int threshold;
        //负载因子
        final float loadFactor;
    

    2.构造函数

    构造函数可以自定义初始化容量和负载因子的大小,其中初始化容量会自动提升为2的n幂次(n>31),提升容量的算法tableSizeFor写的很好

    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);
        }
    
    /**
         * Returns a power of two size for the given target capacity.
         */
        static final int tableSizeFor(int cap) {
            int n = cap - 1;
            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;
        }
    

    n |= n>>>1的作用是将n的二进制中第一个1和这个1的右边一位按位与,结果赋值给n;也就是n的非0最高位起,前两位变成了1;
    n |= n>>>2的作用是将n的二进制中非0最高位中的前两位和n按位与,结果赋值给n;也就是n的非0最高位起,前四位变成1了;
    以此类推...
    n的非0最高位到最低位都是1,然后return n+1,得到最接近的2的幂次

    3.新增方法

    再来看下关键的put方法,put方法会设置map中key对应的value,如果该key存在,覆盖旧值,并将其返回;

    /**
         * 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);
        }
    

    hash()将key的高16位拿来参与运算,这样做的目的是减少哈希碰撞,因为只取低位的值作为hash,碰撞很严重;参考:JDK 源码中 HashMap 的 hash 方法原理是什么?

        static final int hash(Object key) {
            int h;
            return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
        }
    

    putVal()方法是新增节点的实际方法

        /**
         * 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
         */
        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) //空map,按照默认大小或指定参数开辟空间,lazy init
                n = (tab = resize()).length;
            if ((p = tab[i = (n - 1) & hash]) == null) // key的hash值在map中的value如果是空的,
                tab[i] = newNode(hash, key, value, null);//赋新值
            else { //hash位置已存在旧值
                Node<K,V> e; K k;
                if (p.hash == hash &&
                    ((k = p.key) == key || (key != null && key.equals(k)))) //p的值是hash对应的旧的node节点,如果p的key和参数key值相等的话,e = p
                    e = p;
                else if (p instanceof TreeNode) //旧值的key和参数key不相等,并且如果p是树节点,将新的KV放入树里
                    e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
                else {//旧值的key和参数key不相等
                    for (int binCount = 0; ; ++binCount) {
                        if ((e = p.next) == null) { //当前节点是不是链表的尾节点
                            p.next = newNode(hash, key, value, null); //在尾节点新增参数KV
                            if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st //在链表的尾节点判断链表节点的个数是否满足树化的条件
                                treeifyBin(tab, hash);
                            break;
                        }
                        if (e.hash == hash &&
                            ((k = e.key) == key || (key != null && key.equals(k))))//在链表中找和参数key同值或相等的key
                            break;
                        p = e;
                    }
                }
                if (e != null) { // existing mapping for key //e这个节点的作用就是保存以前存在的旧值,用来return
                    V oldValue = e.value;
                    if (!onlyIfAbsent || oldValue == null)
                        e.value = value;
                    afterNodeAccess(e);
                    return oldValue;
                }
            }
            ++modCount;
            if (++size > threshold)//当前节点的个数是否大于阈值
                resize(); //翻倍扩容数组
            afterNodeInsertion(evict);//给LinkedHashMap留的坑位,便于插入节点后处理
            return null;
        }
    

    现在putTreeVal()、treeifyBin()这两个树化操作还有待分析

    4.resize()扩容方法

    /**
         * 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;
            int oldThr = threshold;
            int newCap, newThr = 0; //定义新表容量、新表的阀值
            if (oldCap > 0) { //旧表的容量大于0
                if (oldCap >= MAXIMUM_CAPACITY) { //旧表容量阀值超过定义的最大容量(2的30次方)
                    threshold = Integer.MAX_VALUE; //不扩容,但是调整容量阀值为整形的最大值
                    return oldTab;
                }
                else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY && //原来容量小于1<<30时,新容量为原来容量的二倍
                         oldCap >= DEFAULT_INITIAL_CAPACITY)
                    newThr = oldThr << 1; // double threshold
            }
            else if (oldThr > 0) // initial capacity was placed in threshold //旧表容量小于等于0并且旧表阀值大于0
                newCap = oldThr; //新表容量等于旧表阀值
            else {               // zero initial threshold signifies using defaults //旧表容量和阀值都小于等于0
                newCap = DEFAULT_INITIAL_CAPACITY; //无参初始化,默认容量16,阀值12
                newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
            }
            if (newThr == 0) { //如果没翻倍扩容,也没走默认容量,例如:oldCap<<29,newCap为1<<30时
                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) { //遍历旧数组
                    Node<K,V> e;
                    if ((e = oldTab[j]) != null) { //数组元素不为空
                        oldTab[j] = null;
                        if (e.next == null) //数组元素不是链表,也不是树形结构,即没有hash冲突,只有一个元素的时候
                            newTab[e.hash & (newCap - 1)] = e; //rehash到新表
                        else if (e instanceof TreeNode) //红黑树的mapping
                            ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                        else { // preserve order //维持原来顺序,并且是两条链表(loHead,hiHead),分别进行尾插法
                            Node<K,V> loHead = null, loTail = null;//low位置链表,位置=原表位置
                            Node<K,V> hiHead = null, hiTail = null;//high位置链表,位置=原表位置+原表大小
                            Node<K,V> next;
                            do {
                                next = e.next;
                                if ((e.hash & oldCap) == 0) {//rehash落在和原表相同的位置的元素
                                    if (loTail == null)
                                        loHead = e;
                                    else
                                        loTail.next = e;
                                    loTail = e;
                                }
                                else {//rehash落在后半部分的元素
                                    if (hiTail == null)
                                        hiHead = e;
                                    else
                                        hiTail.next = e;
                                    hiTail = e;
                                }
                            } while ((e = next) != null);
                            if (loTail != null) {
                                loTail.next = null;
                                newTab[j] = loHead; //前一半
                            }
                            if (hiTail != null) {
                                hiTail.next = null;
                                newTab[j + oldCap] = hiHead; //后一半
                            }
                        }
                    }
                }
            }
            return newTab;
        }
    
    • Q:为什么最大容量是2的30次方?
    • A:如果是32位整型的最大值(2的31次方-1),就不能构成翻倍扩容
    • Q:会造成Java1.7的多线程死循环问题吗?
    • A:不会,因为都是尾插法,不会形成环形结构
      假设原数组大小为4,阀值也是4。四个冲突的元素链表如下:

      扩容到8的时候,重新hash后得到两个链表:

    5.根据key获取value

    获取value主要是get()方法

        public V get(Object key) {
            Node<K,V> e;
            return (e = getNode(hash(key), key)) == null ? null : e.value;
        }
    
        /**
         * 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) {//根据hash与数组按位与得到第一个节点
                if (first.hash == hash && // always check first node //判断第一个节点与key是否相等或值相等
                    ((k = first.key) == key || (key != null && key.equals(k))))
                    return first;
                if ((e = first.next) != null) {
                    if (first instanceof TreeNode)
                        return ((TreeNode<K,V>)first).getTreeNode(hash, key);//在树中获取节点
                    do {//在链表中获取节点
                        if (e.hash == hash &&
                            ((k = e.key) == key || (key != null && key.equals(k))))
                            return e;
                    } while ((e = e.next) != null);
                }
            }
            return null;
        }
    

    在树中获取节点getTreeNode()还有待分析

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  • 原文地址:https://www.cnblogs.com/boboshenqi/p/10418238.html
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