源码分析--HashMap（JDK1.8）

　　在jdk1.8中对hashmap的底层实现做了修改。本篇对hashmap源码从核心成员变量到常用方法进行分析。

　　hashmap数据结构如下：

　

　　先看成员变量：

　　1、底层存放数据的是node<k,v>[]数组，数组初始化大小为16。

/**
     * the default initial capacity - must be a power of two.
     */
    static final int default_initial_capacity = 1 << 4; // aka 16

　　2、node<k,v>[]数组最大容量

/**
     * the maximum capacity, used if a higher value is implicitly specified
     * by either of the constructors with arguments.
     * must be a power of two <= 1<<30.
     */
    static final int maximum_capacity = 1 << 30;

　　3、负载因子0.75。也就是如果默认初始化，hashmap在size = 16*0.75 = 12时，进行扩容。

/**
     * the load factor used when none specified in constructor.
     */
    static final float default_load_factor = 0.75f;

　　4、将链表转化为红黑数的阀值。

 /**
     * the bin count threshold for using a tree rather than list for a
     * bin.  bins are converted to trees when adding an element to a
     * bin with at least this many nodes. the value must be greater
     * than 2 and should be at least 8 to mesh with assumptions in
     * tree removal about conversion back to plain bins upon
     * shrinkage.
     */
    static final int treeify_threshold = 8;

　　5、红黑树节点转换为链表的阀值

/**
     * the bin count threshold for untreeifying a (split) bin during a
     * resize operation. should be less than treeify_threshold, and at
     * most 6 to mesh with shrinkage detection under removal.
     */
    static final int untreeify_threshold = 6;

　　6、转红黑树时，table的最小长度

/**
     * the smallest table capacity for which bins may be treeified.
     * (otherwise the table is resized if too many nodes in a bin.)
     * should be at least 4 * treeify_threshold to avoid conflicts
     * between resizing and treeification thresholds.
     */
    static final int min_treeify_capacity = 64;

 

介绍一下hashmap用hash值定位数组index的过程：

//hahsmap中的静态方法
static final int hash(object key) {
        int h;
        return (key == null) ? 0 : (h = key.hashcode()) ^ (h >>> 16);
    }

//定位计算
int index = (table.length - 1) & hash     

• 先得到key的hashcode值
• 再将hashcode值与hashcode无符号右移16位的值进行按位异或运算。得到hash值
• 将(table.length - 1) 与 hash值进行与运算。定位数组index

给一个长度为16的数组，以"testhash"为key，进行定位的过程实例：

hashmap中node就是放入的数据节点，代码定义为：

  static class node<k,v> implements map.entry<k,v> {
        final int hash;
        final k key;
        v value;
        node<k,v> next;
｝

　　node节点保存key的hash值和k--v，借助next可实现链表。

 

红黑树封装为treenode节点：

static final class treenode<k,v> extends linkedhashmap.entry<k,v> {
        treenode<k,v> parent;  // red-black tree links
        treenode<k,v> left;
        treenode<k,v> right;
        treenode<k,v> prev;    // needed to unlink next upon deletion
        boolean red;
        treenode(int hash, k key, v val, node<k,v> next) {
            super(hash, key, val, next);
        }

 

介绍get()方法：

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) {
            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)
                    return ((treenode<k,v>)first).gettreenode(hash, key);
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

• index定位，得到该索引上的node节点，赋值给first
• 对first节点进行判断，如果是要找的元素，直接返回
• first节点的next不为空，继续找
• 如果first节点是红黑树，调用gettreenode()获取值。
• 不是红黑树，只能是链表。从头遍历，找到就返回。

　　上面对于红黑树取值的gettreenode()方法，看一下红黑树的遍历做法：

final treenode<k,v> find(int h, object k, class<?> kc) {
            treenode<k,v> p = this;
            do {
                int ph, dir; k pk;
                treenode<k,v> pl = p.left, pr = p.right, q;
                if ((ph = p.hash) > h)
                    p = pl;
                else if (ph < h)
                    p = pr;
                else if ((pk = p.key) == k || (k != null && k.equals(pk)))
                    return p;
                else if (pl == null)
                    p = pr;
                else if (pr == null)
                    p = pl;
                else if ((kc != null ||
                          (kc = comparableclassfor(k)) != null) &&
                         (dir = comparecomparables(kc, k, pk)) != 0)
                    p = (dir < 0) ? pl : pr;
                else if ((q = pr.find(h, k, kc)) != null)
                    return q;
                else
                    p = pl;
            } while (p != null);
            return null;
        }

• 从do-while循环里的第一个if开始。如果当前节点的hash比传入的hash大，往p节点的左边遍历
• 如果当前节点的hash比传入的hash小，往p节点的右边遍历
• 如果key值相同，就找到节点了。返回
• 左节点为空，转到右边遍历
• 右节点为空，转到左边
• 如果传入key实现了comparable接口。就将传入key与p节点key进行比较，根据比较结果选择向左或向右遍历
• 没有实现接口，直接向右遍历，找到就返回
• 没找到，向左遍历

 

介绍put()方法：

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);
        return null;
    }

• table为null，初始化
• 定位到数组index，若该位置为空，直接放
• 如果该位置上不为空，且hash和key与传入的值相同，说明key重复，直接将该节点赋值给e，结束循环
• 如果该index上的节点是红黑树，调用puttreeval()方法
• 不是红黑树，只能是链表，遍历整个链表
• 找到最后一个节点，在这个节点后面以k--v新增一个节点。
• 判断链表长度，bincount达到7，也就是长度达到8。转为红黑树。
• 遍历过程中，如果找到了相同key，就跳出循环。
• 如果e不为空，说明遍历结束后存在key重复的节点。做值覆盖
• 扩容判断

分析红黑树插入方法puttreeval():

final treenode<k,v> puttreeval(hashmap<k,v> map, node<k,v>[] tab,
                                       int h, k k, v v) {
            class<?> kc = null;
            boolean searched = false;
            treenode<k,v> root = (parent != null) ? root() : this;
            for (treenode<k,v> p = root;;) {
                int dir, ph; k pk;
                if ((ph = p.hash) > h)
                    dir = -1;
                else if (ph < h)
                    dir = 1;
                else if ((pk = p.key) == k || (k != null && k.equals(pk)))
                    return p;
                else if ((kc == null &&
                          (kc = comparableclassfor(k)) == null) ||
                         (dir = comparecomparables(kc, k, pk)) == 0) {
                    if (!searched) {
                        treenode<k,v> q, ch;
                        searched = true;
                        if (((ch = p.left) != null &&
                             (q = ch.find(h, k, kc)) != null) ||
                            ((ch = p.right) != null &&
                             (q = ch.find(h, k, kc)) != null))
                            return q;
                    }
                    dir = tiebreakorder(k, pk);
                }

                treenode<k,v> xp = p;
                if ((p = (dir <= 0) ? p.left : p.right) == null) {
                    node<k,v> xpn = xp.next;
                    treenode<k,v> x = map.newtreenode(h, k, v, xpn);
                    if (dir <= 0)
                        xp.left = x;
                    else
                        xp.right = x;
                    xp.next = x;
                    x.parent = x.prev = xp;
                    if (xpn != null)
                        ((treenode<k,v>)xpn).prev = x;
                    moveroottofront(tab, balanceinsertion(root, x));
                    return null;
                }
            }
        }

• 查找root根节点
• 从root节点开始遍历
• 如果当前节点p的hash大于传入的hash值，记dir为-1，代表向左遍历。
• 小于，记1，代表向右遍历
• 如果key相同，直接返回
• 如果key所属的类实现comparable接口，或者key相等。先从p的左节点、右节点分别调用find()，找到就返回。
• 没找到，比较p和传入的key值，结果记为dir
• 根据dir选择向左或向右遍历
• 依次遍历，直到为null，表示达到最后一个节点，插入新节点
• 调整位置

 

分析hashmap扩容方法：

final node<k,v>[] resize() {
        node<k,v>[] oldtab = table;
        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;
            }
            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);
        }
        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;
                    if (e.next == null)
                        newtab[e.hash & (newcap - 1)] = e;
                    else if (e instanceof treenode)
                        ((treenode<k,v>)e).split(this, newtab, j, oldcap);
                    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;
                            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;
    }

• 通过一系列判断，确认新table的容量
• 构造一个新容量的node数组，赋值给table
• 遍历旧table数组
• 如果节点是单节点，直接定位到新数组对应的index位置下
• 如果是红黑树，调用split方法
• 遍历链表。
• 如果e的hash值与老数组容量取与运算，值为0。索引位置不变
• 如果e的hash值与老数组容量取与运算，值为1。这在新数组中索引的位置为老数组索引 + 老数组容量。
• 链表放置

 

简要分析多线程下hashmap死循环问题：

　　jdk1.7hashmap扩容时，对于链表位置变化，采用头插法进行操作。多线程下容易形成环形链表，造成死循环。

　　jdk1.8时，会对于链表hash值与容量的计算结果。分成两部分，并改为插入到链表尾部。1.8以后不会再有死循环问题，只是有可能重复放置导致数据丢失。hashmap本身线程不安全的特性并没有改变。