【HashMap1.8源码】十分钟带你深入HashMap1.8源码逐行解析

HashMap1.8源码

四个点核心点

• 初始化
• PUT
• 扩容
• GET

初始化

Node结构

transient Node<K,V>[] table;

初始化时为空的Node数组

    static class Node<K,V> implements Map.Entry<K,V> {
        final int hash;
        final K key;
        V value;
        Node<K,V> next;

        Node(int hash, K key, V value, Node<K,V> next) {
            this.hash = hash;
            this.key = key;
            this.value = value;
            this.next = next;
        }

Treenode结构

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);
        }

        /**
         * Returns root of tree containing this node.
         */
        final TreeNode<K,V> root() {
            for (TreeNode<K,V> r = this, p;;) {
                if ((p = r.parent) == null)
                    return r;
                r = p;
            }
        }
    
        static class Entry<K,V> extends HashMap.Node<K,V> {
        Entry<K,V> before, after;
        Entry(int hash, K key, V value, Node<K,V> next) {
            super(hash, key, value, next);
        }
    }

四个构造方法

initialCapacity：初始容量，默认是tableSizeFor(initialCapacity)，根据传参找一个大于该数的2次幂数，比如定义是10，则初始化是16

loadFactor：负载因子，this.loadFactor = DEFAULT_LOAD_FACTOR;默认是0.75，0.75x16 = 12 ，即大于12则扩容if (++size > threshold)

// 构造一：传入初始容量，负载因子    
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);
    }    
	
// 构造二：传入初始容量
	public HashMap(int initialCapacity) {
        this(initialCapacity, DEFAULT_LOAD_FACTOR);
    }

// 构造三：无参构造
    /**
     * 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
    }
// 构造四：传入Map子类
    public HashMap(Map<? extends K, ? extends V> m) {
        this.loadFactor = DEFAULT_LOAD_FACTOR;
        putMapEntries(m, false);
    }

空参关键默认容量是16

static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16

下面才是在resize()方法里创建node数组

        Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];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];

tableSizeFor(initialCapacity)

源码方法

返回给定目标容量的 2 大小的幂

    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;
    }

演算过程

cap = 10 
n = 10 - 1 = 9
9 // 0000 1001
n >>> 1 // 0000 0100
n |= n // 0000 1101  ->13
同理n=13
n >>> 2 // 0000 0011
n |= n // 0000 1111 -> 15    


Integer.highestOneBit(10)); // 8
tableSizeFor(10)); // 16

为什么要2次幂？

n = tab.length

tab[i = (n - 1) & hash]

16: 0001 0000
15:	0000 1111
khash：0000 1010

& 与运算，位上都为1则为1
index：0000 1111

综上，使最高位后面位数都为1，使在与hashcode计算时，更好的散列在容量范围之内

PUT方法

参数一：key

参数二：值

其底层是putVal(hash(key), key, value, false, true);

对应：key的hash值，key，值，onlyIfAbsent，evict

    public V put(K key, V value) {
        return putVal(hash(key), key, value, false, true);
    }

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)
            // 从这里看出，在new时并不会创建tab，而是在put的时候判断是否为空，去resize（）
            n = (tab = resize()).length;
    	// tab[i = (n - 1) & hash] n为什么要2次幂可以看出
        if ((p = tab[i = (n - 1) & hash]) == null)
            // 没有节点则创建一个node节点
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            // 否则hash冲突，看key值是不是一样，是则把p暂存在e，p和e就是已经存在的节点
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            // 这里可以看出在节点里有node和treenode两个类型
            else if (p instanceof TreeNode)
                // 如果是一颗树则插入树上
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                // 判断是一个链表而不是树
                for (int binCount = 0; ; ++binCount) {
                    // 当bincount为0是，此时链表上有一个节点
                    if ((e = p.next) == null) {
                        // 创建节点
                        p.next = newNode(hash, key, value, null);
                        //     static final int TREEIFY_THRESHOLD = 8;
                        // 也就是当bincount为7时为true，即当前有8个节点已经在链表上
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            // 当我创建第9个node时转树
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            // 如果是有hash冲突并且key一样，则把新值覆盖到老值上
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                // 如果onlyIfAbsent为fasle则覆盖老值,默认是false，putIfAbsent方法传入是true
                if (!onlyIfAbsent || oldValue == null)
                    // 把put的值覆盖老值
                    e.value = value;
                // 该方法是linkhashmap操作，空方法
                afterNodeAccess(e);
                // put方法会返回老值
                return oldValue;
            }
        }
        ++modCount;
    // 当当前所有的节点大于threshold时，选择resize()扩容
    //  newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
    // threshold = newThr; = 0.75x16 = 12
    // 也就是第一次有12个节点时就会选择扩容，第二次根据
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

扩容

resize()

cap是容量

threshold是阈值

    /**
     * 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;
        // 第一次是0，第二次是oldTab.length
        int oldCap = (oldTab == null) ? 0 : oldTab.length;
        // 第一次是12
        int oldThr = threshold;
        int newCap, newThr = 0;
        if (oldCap > 0) {
            // 最大阈值是2的32次
            if (oldCap >= MAXIMUM_CAPACITY) {
                threshold = Integer.MAX_VALUE;
                return oldTab;
            }
            // 否则容量x2，当前x2，所以永远是2的次幂
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                // 右移一位，当前x2
                newThr = oldThr << 1; // double threshold
        }
        else if (oldThr > 0) // initial capacity was placed in threshold
            // 初始化容量大于0
            newCap = oldThr;
        else {               // zero initial threshold signifies using defaults
            // 默认初始化的值newCap = 16，newThr = 0.75x16 =12
            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;
        // 存在着有节点的数组，从老数组转移到新数组上
        if (oldTab != null) {
            // 遍历所有的节点
            for (int j = 0; j < oldCap; ++j) {
                Node<K,V> e;
                if ((e = oldTab[j]) != null) {
                    oldTab[j] = null;
                    // 只有一个节点的情况下
                    if (e.next == null)
                        // 转移方法：当前的hash与容量进行与运算得到索引位置，
                        // 也就是为什么要2次幂，因为这样得到的结果只有两个情况
                        // 新下标 = 老下标
                        // 新下标 = 老下标 + 老容量
                        newTab[e.hash & (newCap - 1)] = e;
                    else if (e instanceof TreeNode)
                        // 红黑树转移新数组上
                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                    else { // preserve order
                        // 链表转移新数组上
                        Node<K,V> loHead = null, loTail = null;
                        Node<K,V> hiHead = null, hiTail = null;
                        Node<K,V> next;
                        do {
                            next = e.next;
                            // 判断他的最高位是0还是1，决定了是低位还是高位
                            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);
                        // 低位的放在低位上老下标
                        if (loTail != null) {
                            loTail.next = null;
                            newTab[j] = loHead;
                        }
                        // 高位放在老下标+容量上
                        if (hiTail != null) {
                            hiTail.next = null;
                            newTab[j + oldCap] = hiHead;
                        }
                    }
                }
            }
        }
        return newTab;
    }

转移新数组后，hash值散列问题

khash: 0000 1010
扩容前：
16: 0001 0000
15: 0000 1111

15 0000 1111	0000 1111	
kh 0000 1010	0101 0101
&
i  0000 1010    0000 0101

扩容后：
32: 0010 0000
31: 0001 1111   0001 1111
kh  0000 1010	0101 0101
& 
	0000 1010  --- 新下标 = 老下标
	0001 0101  --- 新下标 = 0000 0101（扩容前hash） + 0001 0000（16扩容前长度） = 老下标 + 老长度
	关键看最高位是0还是1，决定了新下标的位置

treeifyBin()

链表转红黑树

 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
                            // 关键入口 当前有9个节点，也就是节点大于8
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }

    /**
     * Replaces all linked nodes in bin at index for given hash unless
     * table is too small, in which case resizes instead.
     */
//替换给定哈希的 bin at 索引中的所有链接节点，除非表太小，在这种情况下会调整大小。
    final void treeifyBin(Node<K,V>[] tab, int hash) {
        int n, index; Node<K,V> e;
        // static final int MIN_TREEIFY_CAPACITY = 64;
        // 如果数组的长度小于64，还是会选择扩容，而不是转红黑树
        if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
            resize();
        else if ((e = tab[index = (n - 1) & hash]) != null) {
            TreeNode<K,V> hd = null, tl = null;
            do {
                // 把node节点转treenode
                TreeNode<K,V> p = replacementTreeNode(e, null);
                if (tl == null)
                    hd = p;
                else {
                    // 双向链表
                    p.prev = tl;
                    tl.next = p;
                }
                tl = p;
            } while ((e = e.next) != null);
            if ((tab[index] = hd) != null)
                // 把当前的链表转红黑树
                hd.treeify(tab);
        }
    }

 TreeNode<K,V> replacementTreeNode(Node<K,V> p, Node<K,V> next) {
        return new TreeNode<>(p.hash, p.key, p.value, next);
    }

split

红黑树转链表

else if (e instanceof TreeNode)
                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);

        /**
         * Splits nodes in a tree bin into lower and upper tree bins,
         * or untreeifies if now too small. Called only from resize;
         * see above discussion about split bits and indices.
         *
         * @param map the map
         * @param tab the table for recording bin heads
         * @param index the index of the table being split
         * @param bit the bit of hash to split on
         */
//将树箱中的节点拆分为下部和上部树箱，如果现在太小，则取消树化。仅从调整大小调用;请参阅上面关于拆分位和索引的讨论。参数： map – 地图选项卡 – 用于记录垃圾箱头索引的表 – 被拆分的表的索引 位 – 要拆分的哈希位
        final void split(HashMap<K,V> map, Node<K,V>[] tab, int index, int bit) {
            TreeNode<K,V> b = this;
            // Relink into lo and hi lists, preserving order
            TreeNode<K,V> loHead = null, loTail = null;
            TreeNode<K,V> hiHead = null, hiTail = null;
            int lc = 0, hc = 0;
            for (TreeNode<K,V> e = b, next; e != null; e = next) {
                next = (TreeNode<K,V>)e.next;
                e.next = null;
                // 同样要判断高低位，拆分两个
                if ((e.hash & bit) == 0) {
                    if ((e.prev = loTail) == null)
                        loHead = e;
                    else
                        loTail.next = e;
                    loTail = e;
                    ++lc;
                }
                else {
                    if ((e.prev = hiTail) == null)
                        hiHead = e;
                    else
                        hiTail.next = e;
                    hiTail = e;
                    ++hc;
                }
            }
			// 如果存在低位的
            if (loHead != null) {
                // UNTREEIFY_THRESHOLD = 6
                // 并且低位的节点<=6，才会把树转为链表
                if (lc <= UNTREEIFY_THRESHOLD)
                    // 把头节点转入转为链表
                    tab[index] = loHead.untreeify(map);
                else {
                    tab[index] = loHead;
                    // 如果低位节点>6,并且没有高位的节点，就直接把树转移过去
                    if (hiHead != null) // (else is already treeified)
                        loHead.treeify(tab);
                }
            }
            // 判断高位节点
            if (hiHead != null) {
                // <=6
                if (hc <= UNTREEIFY_THRESHOLD)
                    tab[index + bit] = hiHead.untreeify(map);
                else {
                    // 高位的在老下标+oldCap
                    // oldCap = bit
                    tab[index + bit] = hiHead;
                    if (loHead != null)
                        hiHead.treeify(tab);
                }
            }
        }

总结

链表转红黑树：条件是大于8并且数组长度大于64，散列在新数组上不一定在同一个链表上，根据hash高低计算新坐标，拆分成两个链表

红黑树转链表：条件是某一高低位上的节点小于等于6时

putIfAbsent()

当key则存在则不覆盖文章来源地址https://www.toymoban.com/news/detail-617296.html

    public V putIfAbsent(K key, V value) {
        return putVal(hash(key), key, value, true, true);
    }

putTreeVal()

            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);

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;
                }
            }
        }

    /**
     * Replaces all linked nodes in bin at index for given hash unless
     * table is too small, in which case resizes instead.
     */
//替换给定哈希的 bin at 索引中的所有链接节点，除非表太小，在这种情况下会调整大小。
    final void treeifyBin(Node<K,V>[] tab, int hash) {
        int n, index; Node<K,V> e;
        if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
            resize();
        else if ((e = tab[index = (n - 1) & hash]) != null) {
            TreeNode<K,V> hd = null, tl = null;
            do {
                TreeNode<K,V> p = replacementTreeNode(e, null);
                if (tl == null)
                    hd = p;
                else {
                    p.prev = tl;
                    tl.next = p;
                }
                tl = p;
            } while ((e = e.next) != null);
            if ((tab[index] = hd) != null)
                hd.treeify(tab);
        }
    }

GET方法

    public V get(Object key) {
        Node<K,V> e;
        return (e = getNode(hash(key), key)) == null ? null : e.value;
    }

    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;
    }

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