HashMap源码解析（一）-优快云博客

HashMap<K,V>类介绍

HashMap是散列结构，这种结构是支持快速查找的。通过Key计算哈希码，通过哈希码定位到具体的Value(当然具体过程不会这么简单)。在JDK8中HashMap进行了改进，引入了红黑树。JDK8中HashMap是数组+链表+红黑树的复合数据结构。注意：这一篇文章我们先分析HashMap中和红黑树操作无关的部分。

HashMap结构

HashMap中相关字段

/**
 * The default initial capacity - MUST be a power of two.
 */
static final int DEFAULT_INITIAL_CAPACITY = 1 << 4; // aka 16

/**
 * 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;

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

/**
 * 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;

/**
 * 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;

/**
 * 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;

/* ---------------- Fields -------------- */

/**
 * The table, initialized on first use, and resized as
 * necessary. When allocated, length is always a power of two.
 * (We also tolerate length zero in some operations to allow
 * bootstrapping mechanics that are currently not needed.)
 */
transient Node<K,V>[] table;

/**
 * Holds cached entrySet(). Note that AbstractMap fields are used
 * for keySet() and values().
 */
transient Set<Map.Entry<K,V>> entrySet;

/**
 * The number of key-value mappings contained in this map.
 */
transient int size;

/**
 * The number of times this HashMap has been structurally modified
 * Structural modifications are those that change the number of mappings in
 * the HashMap or otherwise modify its internal structure (e.g.,
 * rehash).  This field is used to make iterators on Collection-views of
 * the HashMap fail-fast.  (See ConcurrentModificationException).
 */
transient int modCount;

/**
 * The next size value at which to resize (capacity * load factor).
 *
 * @serial
 */
// (The javadoc description is true upon serialization.
// Additionally, if the table array has not been allocated, this
// field holds the initial array capacity, or zero signifying
// DEFAULT_INITIAL_CAPACITY.)
int threshold;

/**
 * The load factor for the hash table.
 *
 * @serial
 */
final float loadFactor;
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从上面注释我们知道：

DEFAULT_INITIAL_CAPACITY这个字段表示默认table数组的长度，默认是16。
TREEIFY_THRESHOLD表示桶中链表长度达到8个时，会尝试转化为红黑树，后面会介绍。
threshold计算出来的阈值，如果HashMap中存储的元素超过这个阈值，会通过resize进行扩容。
loadFactor加载因子，默认是0.75，我们可以在构造函数中进行动态调整。

HashMap中链表节点Node结构

/**
 * Basic hash bin node, used for most entries.  (See below for
 * TreeNode subclass, and in LinkedHashMap for its Entry subclass.)
 */
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;
    }

    public final K getKey()        { return key; }
    public final V getValue()      { return value; }
    public final String toString() { return key + "=" + value; }

    public final int hashCode() {
        return Objects.hashCode(key) ^ Objects.hashCode(value);
    }

    public final V setValue(V newValue) {
        V oldValue = value;
        value = newValue;
        return oldValue;
    }

    public final boolean equals(Object o) {
        if (o == this)
            return true;
        if (o instanceof Map.Entry) {
            Map.Entry<?,?> e = (Map.Entry<?,?>)o;
            if (Objects.equals(key, e.getKey()) &&
                Objects.equals(value, e.getValue()))
                return true;
        }
        return false;
    }
}
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从源码中我们看到Node节点的结构还是很清晰的，hash,Key,Value和next链接。所以在HashMap中链表只是单链表，LinkedList中的Node是双向链表。

HashMap相关构造函数

/**
 * 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);
}

/**
 * Constructs an empty <tt>HashMap</tt> with the specified initial
 * capacity and the default load factor (0.75).
 *
 * @param  initialCapacity the initial capacity.
 * @throws IllegalArgumentException if the initial capacity is negative.
 */
public HashMap(int initialCapacity) {
    this(initialCapacity, DEFAULT_LOAD_FACTOR);
}

/**
 * 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
}

/**
 * Constructs a new <tt>HashMap</tt> with the same mappings as the
 * specified <tt>Map</tt>.  The <tt>HashMap</tt> is created with
 * default load factor (0.75) and an initial capacity sufficient to
 * hold the mappings in the specified <tt>Map</tt>.
 *
 * @param   m the map whose mappings are to be placed in this map
 * @throws  NullPointerException if the specified map is null
 */
public HashMap(Map<? extends K, ? extends V> m) {
    this.loadFactor = DEFAULT_LOAD_FACTOR;
    putMapEntries(m, false);
}
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HashMap默认都会设置加载因子，我们可以设置加载因子。如果大于0.75，那么Map的利用率是提升(阈值变大，扩容延后了)。这样可能会导致Map发生碰撞的几率更高(查找元素会相对慢一些)。如果小于0.75，那么扩容相对频繁写，但是查找元素可能会快一点(Map发生碰撞的几率小了)。0.75是一个平衡，一般无需做修改。
initialCapacity作为参数传递进来时，会通过tableSizeFor方法计算大于等于输入参数的的最小的2的指数幂例如参数是15，那么输出16；参数是22，输出32。保证HashMap数组长度是2的幂次方。
如果参数是Map的话，直接调用putMapEntries插入元素。下面我们来分析这个函数。

putMapEntries分析

/**
 * Implements Map.putAll and Map constructor
 *
 * @param m the map
 * @param evict false when initially constructing this map, else
 * true (relayed to method afterNodeInsertion).
 */
final void putMapEntries(Map<? extends K, ? extends V> m, boolean evict) {
    int s = m.size();
    if (s > 0) {
        if (table == null) { // pre-size
            float ft = ((float)s / loadFactor) + 1.0F;
            int t = ((ft < (float)MAXIMUM_CAPACITY) ?
                     (int)ft : MAXIMUM_CAPACITY);
            if (t > threshold)
                threshold = tableSizeFor(t);
        }
        else if (s > threshold)
            resize();
        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);
        }
    }
}
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先根据map的大小来计算需要多大的散列表。如果table为空，那么将threshold(扩容阈值)设置为2的幂次方。
如果table不为空，如果map大小超过阈值，那么就先扩容，然后循环调用putValue方法插入元素即可。我们从上面putMapEntries方法中看到，在插入节点的时候，都需要hash(key)计算出散列值，我们看下源码：

hash方法

/**
 * 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;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
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这个计算hash的方法设计的非常巧妙，下面我们来详细分析下这个过程：

h=key.hashCode()。通过这一步先计算出Key键值类型自带的哈希函数，返回int类型的散列值。
将hashCode的高位参与运算，重新计算hash值。下面会解释为什么需要h>>>16。

(n - 1) & hash来计算索引

我们知道HashMap要想获得最好性能，就是计算的Hash值尽可能的均匀分布在每一个桶中。理论上取模运算是一种分布很均匀的算法。但是取模运算性能消耗还是比较大的，在JDK8中，对取模运算进行了优化。通过位与运算来替代取模运算。理论公式是：x mod 2^n = x & (2^n - 1)。我们知道HashMap底层数组的长度总是2的n次方，并且取模运算为“h mod table.length”，对应上面的公式，可以得到该运算等同于“h & (table.length - 1)”。这是HashMap在速度上的优化，因为&比%具有更高的效率。

扰动函数

上面我们在介绍hash计算的时候，看到h>>>16这样的操作。为什么key的hashCode还需要使用高16位进行异或操作呢？

我们先假设没有h>>>16这个操作。看看索引位置如何计算的我们假设HashMap桶的长度是默认值16.现在的索引计算如下：

  10100101 11000100 00100101
& 00000000 00000000 00001111
----------------------------------
  00000000 00000000 00000101        //高位全部归零，只保留末四位
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但这时候问题就来了，这样就算我的散列值分布再松散，要是只取最后几位的话，碰撞也会很严重。更要命的是如果散列本身做得不好，分布上成等差数列的漏洞，恰好使最后几个低位呈现规律性重复，那么碰撞就会更加严重。

扰动函数的作用上面提到过，只是使用最后几位的话，碰撞会很严重，严重降低Map性能。如果我们将高16位也加入运算，就可以较好的解决问题。如图：
右位移16位，正好是32bit的一半，自己的高半区和低半区做异或，就是为了混合原始哈希码的高位和低位，以此来加大低位的随机性(减少碰撞)。而且混合后的低位掺杂了高位的部分特征，这样高位的信息也被变相保留下来。

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'
 */
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;
}
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首先判断table是否为null或者size==0，如果还未初始化或者集合为空，那么先resize进行扩容。
使用p = tab[i = (n - 1) & hash]方法定位到当前的桶在table中的索引。如果当前桶中没有Node节点，那么创建新的Node节点放入桶中即可。否则就说明当前桶中已有元素，需要遍历链表。
else分支首先判断第一个Node节点是否是符合条件的，如果是，整个查找过程就结束了。如果不是，判断第一个节点是否是树节点(这个桶是否已经树化)。如果已经转化成红黑树，就调用红黑树的插入操作。否则就是普通的链表遍历操作。如果整个遍历下来，都没有找到符合条件的的Node节点的话，就构造一个Node节点，放入链表尾部即可，当然，如果链表的长度超过8，会调用treeifyBin方法将这个链表转化为红黑树。否则将找到的节点赋值给e即可。
如果e不为空的话，就说明Map中已有符合条件的节点，新的Value值可以根据参数决定是否覆盖旧值。
最后size++，判断节点数量是否超过threshold阈值，超过的话需要扩容。
从源码看出，Map中判断一个元素是否相等是否的是==或者equal()方法。所以我们在自定义类中，需要好好设计equal()方法和hashCode方法，因为这关乎到Map中元素的查找与比较。

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;
    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;
}
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首先根据扩容前的容量oldCap，如果oldCap容量已经到最大值了，那么不进行扩容，只是将阈值设置为Integer.MAX_VALUE。否则就是扩容为原来的2倍。
如果oldCap==0，但是oldThr不为空的时候（因为构造HashMap初始容量被放入阈值），会将容量设置为当前的阈值。
Map的容量和阈值都是0时，是一个空表，Map容量设置为DEFAULT_INITIAL_CAPACITY大小(16)。并计算出阈值。
如果新的阈值为0，就根据新的容量和加载因子计算出新的阈值。
开始遍历老的Map集合，将里面的Node节点重新定位到新的Map集合中。如果桶中只有一个元素，通过newTab[e.hash & (newCap - 1)] = e;计算出该Node在新Map中的索引即可。
否则判断该桶是否已经树化，如果树化，调用树节点的方法进行hash分布。否则就需要将链表数据一个个遍历，重新定位。此处：HashMap的方法设计的非常精妙。通过定义loHead、loTail、hiHead、hiTail来讲一个链表拆分成两个独立的链表。注意：如果e的hash值与老表的容量进行与运算为0,则扩容后的索引位置跟老表的索引位置一样。所以loHead-->loTail组成的链表在新Map中的索引位置和老Map中是一样的。如果e的hash值与老表的容量进行与运算为1,则扩容后的索引位置为:老表的索引位置＋oldCap。
最后将loHead、loTail、hiHead、hiTail组成的两条链表重新定位到新的Map中即可。

扩容代码(e.hash & oldCap)是否为0来定位的问题

扩容代码中，使用e节点的hash值跟oldCap进行位与运算，以此决定将节点分布到原索引位置或者原索引+oldCap位置上，为什么可以这样计算，我们来看例子：
假设老表的容量为16，即oldCap=16，则新表容量为16*2=32，假设节点1的hash值为0000 0000 0000 0000 0000 1111 0000 1010，节点2的hash值为0000 0000 0000 0000 0000 1111 0001 1010，则节点1和节点2在老表的索引位置计算如下图计算1，由于老表的长度限制，节点1和节点2的索引位置只取决于节点hash值的最后4位。再看计算2，计算2为新表的索引计算，可以知道如果两个节点在老表的索引位置相同，则新表的索引位置只取决于节点hash值倒数第5位的值，而此位置的值刚好为老表的容量值16，此时节点在新表的索引位置只有两种情况：原索引位置和原索引+oldCap位置（在此例中即为10和10+16=26）。由于结果只取决于节点hash值的倒数第5位，而此位置的值刚好为老表的容量值16，因此此时新表的索引位置的计算可以替换为计算3，直接使用节点的hash值与老表的容量16进行位于运算，如果结果为0则该节点在新表的索引位置为原索引位置，否则该节点在新表的索引位置为原索引+oldCap位置。

treeifyBin方法

/**
 * Replaces all linked nodes in bin at index for given hash unless
 * table is too small, in which case resizes instead.
 */
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);
    }
}
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这个代码是将链表转化成红黑树的。我们来看主要的逻辑：

如果Map的容量小于MIN_TREEIFY_CAPACITY(64)。是不会讲某一个桶中的链表转化为红黑树的，会对Map进行扩容。这样每一个桶中的链表的长度会减少。
如果Map的容量符合要求了，那就将链表转化成红黑树。

get方法

/**
 * 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;
}

/**
 * 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) {
        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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HashMap是高效的查找数据结构，所以get方法我们有必要好好分析下。

首先计算出key的哈希值，然后使用first = tab[(n - 1) & hash])定位到索引位置。
首先判断第一个Node节点是否是符合要求的，符合要求就直接返回即可。否则判断当前桶的结构是树结构还是链表结构，分别使用相关的方法寻找节点。没有找到就返回null。

remove方法

/**
 * Removes the mapping for the specified key from this map if present.
 *
 * @param  key key whose mapping is to be removed from the map
 * @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 remove(Object key) {
    Node<K,V> e;
    return (e = removeNode(hash(key), key, null, false, true)) == null ?
        null : e.value;
}

/**
 * Implements Map.remove and related methods
 *
 * @param hash hash for key
 * @param key the key
 * @param value the value to match if matchValue, else ignored
 * @param matchValue if true only remove if value is equal
 * @param movable if false do not move other nodes while removing
 * @return the node, or null if none
 */
final Node<K,V> removeNode(int hash, Object key, Object value,
                           boolean matchValue, boolean movable) {
    Node<K,V>[] tab; Node<K,V> p; int n, index;
    if ((tab = table) != null && (n = tab.length) > 0 &&
        (p = tab[index = (n - 1) & hash]) != null) {
        Node<K,V> node = null, e; K k; V v;
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            node = p;
        else if ((e = p.next) != null) {
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            else {
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key ||
                         (key != null && key.equals(k)))) {
                        node = e;
                        break;
                    }
                    p = e;
                } while ((e = e.next) != null);
            }
        }
        if (node != null && (!matchValue || (v = node.value) == value ||
                             (value != null && value.equals(v)))) {
            if (node instanceof TreeNode)
                ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
            else if (node == p)
                tab[index] = node.next;
            else
                p.next = node.next;
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
}
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