java.util.HashMap阅读笔记

HashMap阅读笔记

这是我读的第1个源码, mark一下吧.

强烈推荐 JDK8 HashMap源码行级解析 红黑树操作 史上最全最详细图解_anlian523的博客-CSDN博客. 关于红黑树的部分讲解的很好.

建议如果阅读代码中的笔记, 将下面的代码块拷贝到vs code或idea上. 因为在markdown的代码块中宽度有限制, 会影响观看效果.

/*
 * Copyright (c) 1997, 2017, Oracle and/or its affiliates. All rights reserved.
 * ORACLE PROPRIETARY/CONFIDENTIAL. Use is subject to license terms.
 *
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 */

package java.util;

import java.io.IOException;
import java.io.InvalidObjectException;
import java.io.Serializable;
import java.lang.reflect.ParameterizedType;
import java.lang.reflect.Type;
import java.util.function.BiConsumer;
import java.util.function.BiFunction;
import java.util.function.Consumer;
import java.util.function.Function;
import sun.misc.SharedSecrets;

/**
 * Hash table based implementation of the <tt>Map</tt> interface.  This
 * implementation provides all of the optional map operations, and permits
 * <tt>null</tt> values and the <tt>null</tt> key.  (The <tt>HashMap</tt>       这里说明了允许key和value为 null, 并且是unsynchronized的
 * class is roughly equivalent to <tt>Hashtable</tt>, except that it is         而Hashtable不允许null, 并且是synchronized的
 * unsynchronized and permits nulls.)  This class makes no guarantees as to
 * the order of the map; in particular, it does not guarantee that the order    不保证顺序永远一样. 这一点很容易理解, 一扩容就要重新计算hash % len, 根据这个值来确定顺序
 * will remain constant over time.
 *
 * <p>This implementation provides constant-time performance for the basic
 * operations (<tt>get</tt> and <tt>put</tt>), assuming the hash function
 * disperses the elements properly among the buckets.  Iteration over             bucket为数组中的每个元素
 * collection views requires time proportional to the "capacity" of the
 * <tt>HashMap</tt> instance (the number of buckets) plus its size (the number
 * of key-value mappings).  Thus, it's very important not to set the initial
 * capacity too high (or the load factor too low) if iteration performance is
 * important.
 *
 * <p>An instance of <tt>HashMap</tt> has two parameters that affect its
 * performance: <i>initial capacity</i> and <i>load factor</i>.  The
 * <i>capacity</i> is the number of buckets in the hash table, and the initial      capacity是bucket的数量而不是(key, value)的数量
 * capacity is simply the capacity at the time the hash table is created.  The      hashmap中的键值对(key, value)被称为entry
 * <i>load factor</i> is a measure of how full the hash table is allowed to
 * get before its capacity is automatically increased.  When the number of
 * entries in the hash table exceeds the product of the load factor and the        当entry的数量大于 loadFactor * currentCapacity就扩容到2的下一个次方
 * current capacity, the hash table is <i>rehashed</i> (that is, internal data
 * structures are rebuilt) so that the hash table has approximately twice the
 * number of buckets.
 *
 * <p>As a general rule, the default load factor (.75) offers a good              loadFactor为0.75是空间占用和查找效率的tradeoff
 * tradeoff between time and space costs.  Higher values decrease the
 * space overhead but increase the lookup cost (reflected in most of
 * the operations of the <tt>HashMap</tt> class, including
 * <tt>get</tt> and <tt>put</tt>).  The expected number of entries in
 * the map and its load factor should be taken into account when
 * setting its initial capacity, so as to minimize the number of                    初始size大于entry数量除以loadfactor的时候也不会缩小hashmap大小 
 * rehash operations.  If the initial capacity is greater than the                    这里表示哈希表只会扩容不会缩小
 * maximum number of entries divided by the load factor, no rehash
 * operations will ever occur.
 *
 * <p>If many mappings are to be stored in a <tt>HashMap</tt>
 * instance, creating it with a sufficiently large capacity will allow               
 * the mappings to be stored more efficiently than letting it perform
 * automatic rehashing as needed to grow the table.  Note that using
 * many keys with the same {@code hashCode()} is a sure way to slow
 * down performance of any hash table. To ameliorate impact, when keys             ameliorate == make it perform better
 * are {@link Comparable}, this class may use comparison order among               当很多key的hashcode都相等的时候, 如果key实现了Comparable接口, 
 * keys to help break ties.                                                        就根据这些key的大小关系来改善性能. (事实上就是用大小关系构造红黑树)
 *
 * <p><strong>Note that this implementation is not synchronized.</strong>         hashmap线程不安全
 * If multiple threads access a hash map concurrently, and at least one of        如果多个线程并发访问, 并且至少一个改变hashmap的结构， 就要加锁
 * the threads modifies the map structurally, it <i>must</i> be
 * synchronized externally.  (A structural modification is any operation
 * that adds or deletes one or more mappings; merely changing the value
 * associated with a key that an instance already contains is not a
 * structural modification.)  This is typically accomplished by
 * synchronizing on some object that naturally encapsulates the map.
 *
 * If no such object exists, the map should be "wrapped" using the
 * {@link Collections#synchronizedMap Collections.synchronizedMap}
 * method.  This is best done at creation time, to prevent accidental            要直接在参数中生成, 而不是保留内部集合的引用,防止意外的获得非同步引用. 
 * unsynchronized access to the map:<pre>                                        错误示例:
 *   Map m = Collections.synchronizedMap(new HashMap(...));</pre>                HashMap<?, ?> map = new HashMap<>();
 *                                                                               Map p = Collections.synchronizedMap(map); 意外的使用非同步引用map是危险的
 *
 * <p>The iterators returned by all of this class's "collection view methods"
 * are <i>fail-fast</i>: if the map is structurally modified at any time after     创建迭代器后
 * the iterator is created, in any way except through the iterator's own           任何修改都会引发异常, 除了迭代器自己的remove方法
 * <tt>remove</tt> method, the iterator will throw a
 * {@link ConcurrentModificationException}.  Thus, in the face of concurrent
 * modification, the iterator fails quickly and cleanly, rather than risking
 * arbitrary, non-deterministic behavior at an undetermined time in the
 * future.
 *
 * <p>Note that the fail-fast behavior of an iterator cannot be guaranteed       但是不能依赖快速失败的特点来编写程序. 这样的正确性不能保证
 * as it is, generally speaking, impossible to make any hard guarantees in the
 * presence of unsynchronized concurrent modification.  Fail-fast iterators
 * throw <tt>ConcurrentModificationException</tt> on a best-effort basis.
 * Therefore, it would be wrong to write a program that depended on this
 * exception for its correctness: <i>the fail-fast behavior of iterators
 * should be used only to detect bugs.</i>
 *
 * <p>This class is a member of the
 * <a href="{@docRoot}/../technotes/guides/collections/index.html">
 * Java Collections Framework</a>.
 *
 * @param <K> the type of keys maintained by this map
 * @param <V> the type of mapped values
 *
 * @author  Doug Lea
 * @author  Josh Bloch
 * @author  Arthur van Hoff
 * @author  Neal Gafter
 * @see     Object#hashCode()
 * @see     Collection
 * @see     Map
 * @see     TreeMap
 * @see     Hashtable
 * @since   1.2
 */
public class HashMap<K,V> extends AbstractMap<K,V>
    implements Map<K,V>, Cloneable, Serializable {

    private static final long serialVersionUID = 362498820763181265L;

    /*
     * Implementation notes.
     *
     * This map usually acts as a binned (bucketed) hash table, but    这里的bin推测是每个bucket中的各个entry
     * when bins get too large, they are transformed into bins of
     * TreeNodes, each structured similarly to those in
     * java.util.TreeMap. Most methods try to use normal bins, but     大多数方法都是认为bin是普通的节点, 直接对节点进行操作
     * relay to TreeNode methods when applicable (simply by checking   如果发现是tree bin的节点, 就去调用操作tree bin的函数
     * instanceof a node).  Bins of TreeNodes may be traversed and
     * used like any others, but additionally support faster lookup
     * when overpopulated. However, since the vast majority of bins in
     * normal use are not overpopulated, checking for existence of
     * tree bins may be delayed in the course of table methods.
     *
     * Tree bins (i.e., bins whose elements are all TreeNodes) are            tree bin的红黑树按照hash值排序, (主要是按照hashcode排序)
     * ordered primarily by hashCode, but in the case of ties, if two         我一直搞不懂这个ties是什么意思, 直到看到TreeNode的find方法才明白, ties是指两个对象hash相同的情况
     * elements are of the same "class C implements Comparable<C>",           当两个对象hash值相同的时候, 如果C实现了comparable接口, 可以用这个compareTo方法. 
     * type then their compareTo method is used for ordering. (We             
     * conservatively check generic types via reflection to validate          
     * this -- see method comparableClassFor).  The added complexity         使用反射查看泛型参数是不是实现了Comparable接口 
     * of tree bins is worthwhile in providing worst-case O(log n)            
     * operations when keys either have distinct hashes or are               如果hash不相同, 或者是可排序的, 最坏情况下就是log n的复杂度
     * orderable, Thus, performance degrades gracefully under
     * accidental or malicious usages in which hashCode() methods
     * return values that are poorly distributed, as well as those in
     * which many keys share a hashCode, so long as they are also            如果很多hash相同, 但是他们是comparable的, 性能会下降
     * Comparable. (If neither of these apply, we may waste about a          
     * factor of two in time and space compared to taking no
     * precautions. But the only known cases stem from poor user
     * programming practices that are already so slow that this makes
     * little difference.)
     *
     * Because TreeNodes are about twice the size of regular nodes, we
     * use them only when bins contain enough nodes to warrant use
     * (see TREEIFY_THRESHOLD). And when they become too small (due to
     * removal or resizing) they are converted back to plain bins.  In
     * usages with well-distributed user hashCodes, tree bins are
     * rarely used.  Ideally, under random hashCodes, the frequency of      这里使用泊松分布说明了如果hashcode均匀, 出现tree bin的概率非常小
     * nodes in bins follows a Poisson distribution
     * (http://en.wikipedia.org/wiki/Poisson_distribution) with a
     * parameter of about 0.5 on average for the default resizing
     * threshold of 0.75, although with a large variance because of
     * resizing granularity. Ignoring variance, the expected
     * occurrences of list size k are (exp(-0.5) * pow(0.5, k) /
     * factorial(k)). The first values are:
     *
     * 0:    0.60653066
     * 1:    0.30326533
     * 2:    0.07581633
     * 3:    0.01263606
     * 4:    0.00157952
     * 5:    0.00015795
     * 6:    0.00001316
     * 7:    0.00000094
     * 8:    0.00000006
     * more: less than 1 in ten million
     *
     * The root of a tree bin is normally its first node.  However,
     * sometimes (currently only upon Iterator.remove), the root might
     * be elsewhere, but can be recovered following parent links
     * (method TreeNode.root()).
     *
     * All applicable internal methods accept a hash code as an
     * argument (as normally supplied from a public method), allowing   所有的内部方法都接受一个hash变量来避免重复计算key的hash
     * them to call each other without recomputing user hashCodes.
     * Most internal methods also accept a "tab" argument, that is      大部分内部方法也接受一个tab变量, 大部分时间是当前数组的引用, 
     * normally the current table, but may be a new or old one when     但是会因为resize等操作更新
     * resizing or converting.
     *
     * When bin lists are treeified, split, or untreeified, we keep    树化或非树化之后的遍历顺序不变, 都是next指针维持的顺序
     * them in the same relative access/traversal order (i.e., field
     * Node.next) to better preserve locality, and to slightly
     * simplify handling of splits and traversals that invoke
     * iterator.remove. When using comparators on insertion, to keep a
     * total ordering (or as close as is required here) across
     * rebalancings, we compare classes and identityHashCodes as
     * tie-breakers.
     *
     * The use and transitions among plain vs tree modes is
     * complicated by the existence of subclass LinkedHashMap. See     LinkedHashMap的树化, 非树化方法更复杂
     * below for hook methods defined to be invoked upon insertion,
     * removal and access that allow LinkedHashMap internals to
     * otherwise remain independent of these mechanics. (This also
     * requires that a map instance be passed to some utility methods
     * that may create new nodes.)
     *
     * The concurrent-programming-like SSA-based coding style helps
     * avoid aliasing errors amid all of the twisty pointer operations.
     */

    /**
     * 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;  //并不是链表长度一超过8就变为红黑树. 而是还要满足table.length >= MIN_TREEIFY_CAPACITY才可以. 如果不超过, 就不树化了, 而是resize
                                                //所以hashmap扩容触发条件有2个, 第一个就是size大于了capacity * loadfactor, 第二个就是table.length小于MIN_TREEIFY_CAPACITY, 而且链表长度超过了8
    /**
     * 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> {   //entry的真正实现类
        final int hash;
        final K key;
        V value;
        Node<K,V> next;                                   //指向下一个元素的指针

        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);    //Entry的hashcode是key的hash和value的hash异或得到的
        }

        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;
        }
    }

    /* ---------------- Static utilities -------------- */

    /**
     * Computes key.hashCode() and spreads (XORs) higher bits of hash
     * to lower.  Because the table uses power-of-two masking, sets of   由于table的长度总是2的整数次幂, 所以高位的不同并不会避免碰撞.
     * hashes that vary only in bits above the current mask will         所以要做一个变换使得高位的hashcode能够影响低位的
     * 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         这是速度和可用性的tradeoff. 
     * quality of bit-spreading. Because many common sets of hashes      因为很多类的哈希值已经是被设计为良好分布的了, 所以这个变换并不会使得他们受益很多
     * are already reasonably distributed (so don't benefit from         并且因为有树化的设计, 碰撞了复杂度也不会提高很多.
     * spreading), and because we use trees to handle large sets of      所以没必要用一个复杂的变换. 仅仅用一个不花费太多时间的异或操作即可.
     * 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);   //注意到null的hash为0
    }

    /**
     * Returns x's Class if it is of the form "class C implements
     * Comparable<C>", else null.
     */
    static Class<?> comparableClassFor(Object x) {                     //通过反射获得实现接口Comparable的类的Class对象
        if (x instanceof Comparable) {
            Class<?> c; Type[] ts, as; Type t; ParameterizedType p;
            if ((c = x.getClass()) == String.class) // bypass checks
                return c;
            if ((ts = c.getGenericInterfaces()) != null) {
                for (int i = 0; i < ts.length; ++i) {
                    if (((t = ts[i]) instanceof ParameterizedType) &&
                        ((p = (ParameterizedType)t).getRawType() ==
                         Comparable.class) &&
                        (as = p.getActualTypeArguments()) != null &&
                        as.length == 1 && as[0] == c) // type arg is c
                        return c;
                }
            }
        }
        return null;
    }

    /**
     * Returns k.compareTo(x) if x matches kc (k's screened comparable
     * class), else 0.
     */
    @SuppressWarnings({"rawtypes","unchecked"}) // for cast to Comparable
    static int compareComparables(Class<?> kc, Object k, Object x) {
        return (x == null || x.getClass() != kc ? 0 :
                ((Comparable)k).compareTo(x));
    }

    /**
     * Returns a power of two size for the given target capacity.
     */
    static final int tableSizeFor(int cap) {                 //异或获得大于cap的最小的2的整数次幂
        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;
    }

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

    /**
     * The table, initialized on first use, and resized as
     * necessary. When allocated, length is always a power of two.    长度总是2的整数次幂
     * (We also tolerate length zero in some operations to allow      长度是0也可以
     * 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;                                       //迭代器就是根据这个变量来决定是否
                                                                  //抛出ConcurrentModificationException
    /**
     * The next size value at which to resize (capacity * load factor).
     *
     * @serial
     */
    // (The javadoc description is true upon serialization.                //并不是new完一个hashmap就立即创建table数组. 刚new完hashmap之后table[]数组为null
    // Additionally, if the table array has not been allocated, this       //直到put第一个元素的时候才会通过resize()方法创建table[]数组.
    // field holds the initial array capacity, or zero signifying
    // DEFAULT_INITIAL_CAPACITY.)              //如果table已经被创建了, 那么threshold为capacity * load factor. 超过这个值就扩容
    int threshold;                           //在table还未被创建的时候, 这个值等于即将要创建的table[]数组的长度. 也就是说第一次resize就把table变为大小为threshold的数组.
    /**                                      //如果使用空参构造器, 这个值初始为0, 否则这个值为大于等于capacity的第一个2的整数次幂.capacity为传入的指定大小
     * The load factor for the hash table.
     *
     * @serial
     */
    final float loadFactor;

    /* ---------------- Public operations -------------- */

    /**
     * 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)                            //超过2^30的初始容量也会被变成2^30的容量
            initialCapacity = MAXIMUM_CAPACITY;
        if (loadFactor <= 0 || Float.isNaN(loadFactor))
            throw new IllegalArgumentException("Illegal load factor: " +
                                               loadFactor);
        this.loadFactor = loadFactor;
        this.threshold = tableSizeFor(initialCapacity);                   //刚创建hashmap后, threshold为大于等于初始容量的第一个2的整数次幂
    }                                                                     //table在new完hashmap之后, 插入值之前暂时为空

    /**
     * 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               //threshold = 0
    }

    /**
     * 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) {//用一个map作为构造器参数
        this.loadFactor = DEFAULT_LOAD_FACTOR;       //首先要确定新的map的table的大小. 通过旧的map元素数量, loadFactor算出来. 仍然是2的整数次幂
        putMapEntries(m, false);
    }

    /**
     * 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);
            }
        }
    }

    /**
     * Returns the number of key-value mappings in this map.
     *
     * @return the number of key-value mappings in this map
     */
    public int size() {
        return size;
    }

    /**
     * Returns <tt>true</tt> if this map contains no key-value mappings.
     *
     * @return <tt>true</tt> if this map contains no key-value mappings
     */
    public boolean isEmpty() {
        return size == 0;
    }

    /**
     * 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) {                       //先检查对应桶中第一个是不是, 不是在判断是链表还是树. 是树就调用树的方法, 是链表就从头到尾开始查
        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                 //利用短路逻辑. 而不是直接就比较 key.equals(k)
                ((k = first.key) == key || (key != null && key.equals(k))))     //如果hash值不等, 就肯定不相等.
                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;
    }

    /**
     * Returns <tt>true</tt> if this map contains a mapping for the
     * specified key.
     *
     * @param   key   The key whose presence in this map is to be tested
     * @return <tt>true</tt> if this map contains a mapping for the specified
     * key.
     */
    public boolean containsKey(Object key) {
        return getNode(hash(key), key) != null;
    }

    /**
     * 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);
    }

    /**
     * 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,  //1. 检查hashmap是不是第一次插入值, 如果是, 要调用resize初始化table
                   boolean evict) {                                 //2. 查看对应的桶中第一个元素是不是对应的值, 如果是, 直接修改这个值
        Node<K,V>[] tab; Node<K,V> p; int n, i;                     //3. 查看当前桶是链表还是树, 如果是链表, 依次查找, 找不到就在链表尾部插入一个新的node, 插入之后要检查是不是要树化
        if ((tab = table) == null || (n = tab.length) == 0)         //4. 如果是树, 调用树的put方法putTreeVal.
            n = (tab = resize()).length;                            //5. 如果插入了, 就++size, 再次判断是不是需要resize. 如果是替换, 就返回旧的值
        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))))              //同样先检查第一个元素
                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);  //新元素插入到链表尾部
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st         //binCount是当前链表未插入new出来的节点的的长度, 
                            treeifyBin(tab, hash);                                  //如果只看这一行可能会觉得判断条件是插入新节点
                        break;                                                     //后的链表长度大于等于TREEIFY_THRESHOLD就可以了. 其实不是.
                    }                                                             //这个长度是从第二个节点开始算的. 因为第一个节点我们之前已经判断过了
                    if (e.hash == hash &&                                        //所以是当总链表长度 大于 8 的时候才转换成树的. 如果链表插入元素之后的长度刚好为8, 是不进行转换的
                        ((k = e.key) == key || (key != null && key.equals(k)))) //也可以理解为, 插入元素之前, 链表长度达到8, 并且要插入新元素, 就转化为红黑树
                        break;
                    p = e;
                }
            }
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);  //默认的afterNodeAccess()方法为空函数, 如果有需求, 可以直接改这个函数作为回调, 不得不说HashMap的设计者很贴心
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict); //和afterNodeAccess一样, 也是个空方法, 方便我们增加回调机制
        return null;
    }

    /**
     * Initializes or doubles table size.  If null, allocates in               //第1步. 检查table是不是null, 如果是, 说明是第一次resize,  
     * accord with initial capacity target held in field threshold.            // 第一次的table长度要么是默认的16, 要么是threshold, 然后重新赋值threshold为capacity * loadfactor
     * Otherwise, because we are using power-of-two expansion, the             // 如果不是第一次, 直接新的长度和新的threshold 乘2, 但是要注意, 如果达到了最大值, 就不乘2了, threshold也直接定义为INT_MAX
     * elements from each bin must either stay at same index, or move          // 确定好长度之和就要new新的table了, 然后把原来的node放进去. 首先装第一个节点, 然后判断剩下的
     * with a power of two offset in the new table.                            // 如果是树化的, 调用方法split. 如果是链表, 拆成2部分, 装入两个不同的桶中
     *
     * @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) {                              //如果不是第一次resize, 即table != null, 那么就扩展新的容量为2倍, 新的threshold也扩展为原来的2倍
            if (oldCap >= MAXIMUM_CAPACITY) {          // 如果旧的容量达到最大值, 就不扩容了, 而只是更新threshold为INT_MAX
                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   //如果是第一次resize, 即table为null, 那么第一次table应该设置为threshold大小的数组
            newCap = oldThr;                                               // 如果使用无参构造器, threshold为0, 就设置table大小为16. 
        else {               // zero initial threshold signifies using defaults            
            newCap = DEFAULT_INITIAL_CAPACITY;
            newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
        } 
        if (newThr == 0) {                                            //设置完table的大小之后, 就更新threshold为loadfactor * newCapacity
            float ft = (float)newCap * loadFactor;                    // 如果新threshold和新capacity有一个超过2^30, 就把threshold设置为INT_MAX. 
            newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ? //为了使得继续添加元素也不会引起hashmap再次扩容
                      (int)ft : Integer.MAX_VALUE);
        }
        threshold = newThr;
        @SuppressWarnings({"rawtypes","unchecked"})
        Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];        //下面是根据上面计算出来的newCapacity来新建table[]数组
        table = newTab;
        if (oldTab != null) {                 //如果原来的table为null, 就直接返回新的数组即可. 否则还要将原来的元素迁移到新数组中
            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;             //先把第一个节点放到正确的桶中
                    else if (e instanceof TreeNode)
                        ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);//如果是树状节点, 调用对应的方法
                    else { // preserve order
                        Node<K,V> loHead = null, loTail = null;  
                        Node<K,V> hiHead = null, hiTail = null;
                        Node<K,V> next;
                        do {                                //(e.hash & oldCap) == 0这一个判断很关键. 这个操作将原来的节点分成2部分, 
                            next = e.next;                 //第一部分是模newCap还是原来的值的, 第二个部分是模newCap是原来的值加oldcap的
                            if ((e.hash & oldCap) == 0) { //例如, oldCap = 16, 某个桶内的元素哈希值为 1, 17, 33, 49, 65, 这些值模16都为1
                                if (loTail == null)      //但是扩容之后 newCap = 32, 这些值模32之后就变为了2部分, 一部分模32仍然是1, (1, 33, 65)
                                    loHead = e;         //另一部分模32之后就变为了1+oldCap, 即17. (17, 49). 这两部分在新的table中是要放到不同的桶中的
                                else                   //而(e.hash & oldCap) == 0就是判断这些元素到底要放到哪个桶中.
                                    loTail.next = e;
                                loTail = e;            //假设某个桶中所有元素模oldCap都为a, 遍历过程中loHead和loTail存储模newCap仍为a的元素, 
                            }                          //hiHead和hiTail存储模newCap为a + oldCap的元素, 并且这些元素的相对位置不变
                            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;
    }

    /**
     * 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)      //如果table.length小于MIN_TREEIFY_CAPACITY就不树化, 而是table扩容
            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);        //将链表的所有节点转化为树节点, 结构仍然是链表. 
                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);                                    //将链表树化为红黑树
        }
    }

    /**
     * Copies all of the mappings from the specified map to this map.
     * These mappings will replace any mappings that this map had for
     * any of the keys currently in the specified map.
     *
     * @param m mappings to be stored in this map
     * @throws NullPointerException if the specified map is null
     */
    public void putAll(Map<? extends K, ? extends V> m) {
        putMapEntries(m, true);
    }

    /**
     * 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) {      // 移除元素. 注意到, 移除很多个元素也不会触发resize缩小hashmap大小.
        Node<K,V>[] tab; Node<K,V> p; int n, index;                        //hashmap只能扩容不能缩容
        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 &&                                             //这里的if主要是找到给定key对应的引用node
                ((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;      //如果跳出循环, p就是e的前一个结点. 即p是node的前一个节点. 根据这个来进行删除操作
                            break;
                        }
                        p = e;
                    } while ((e = e.next) != null);
                }
            }                                                                            //至此, 拿到了要删除的节点node
            if (node != null && (!matchValue || (v = node.value) == value ||           //这里的if主要是删除node
                                 (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);//可以定义删除节点后的触发操作. 这里函数体为空
                return node;
            }
        }
        return null;
    }

    /**
     * Removes all of the mappings from this map.
     * The map will be empty after this call returns.
     */
    public void clear() {
        Node<K,V>[] tab;
        modCount++;
        if ((tab = table) != null && size > 0) {
            size = 0;
            for (int i = 0; i < tab.length; ++i)
                tab[i] = null;
        }
    }

    /**
     * Returns <tt>true</tt> if this map maps one or more keys to the
     * specified value.
     *
     * @param value value whose presence in this map is to be tested
     * @return <tt>true</tt> if this map maps one or more keys to the
     *         specified value
     */
    public boolean containsValue(Object value) {
        Node<K,V>[] tab; V v;
        if ((tab = table) != null && size > 0) {
            for (int i = 0; i < tab.length; ++i) {
                for (Node<K,V> e = tab[i]; e != null; e = e.next) {   //注意, TreeNode也有next指针, 可以这样来遍历
                    if ((v = e.value) == value ||
                        (value != null && value.equals(v)))           
                        return true;
                }
            }
        }
        return false;
    }

    /**
     * Returns a {@link Set} view of the keys contained in this map.
     * The set is backed by the map, so changes to the map are
     * reflected in the set, and vice-versa.  If the map is modified
     * while an iteration over the set is in progress (except through
     * the iterator's own <tt>remove</tt> operation), the results of
     * the iteration are undefined.  The set supports element removal,
     * which removes the corresponding mapping from the map, via the
     * <tt>Iterator.remove</tt>, <tt>Set.remove</tt>,
     * <tt>removeAll</tt>, <tt>retainAll</tt>, and <tt>clear</tt>
     * operations.  It does not support the <tt>add</tt> or <tt>addAll</tt>
     * operations.
     *
     * @return a set view of the keys contained in this map
     */
    public Set<K> keySet() {
        Set<K> ks = keySet;
        if (ks == null) {
            ks = new KeySet();
            keySet = ks;
        }
        return ks;
    }
    //写到这里才想起来, 原来Map没有iterator()方法, 不能通过 map =  new HashMap(); map.iterator();来获取迭代器, 所以只能通过先获取KeySet, Values, EntrySet再迭代的方式
    final class KeySet extends AbstractSet<K> {                      //内部类, 指向外围对象的指针为HashMap.this
        public final int size()                 { return size; }
        public final void clear()               { HashMap.this.clear(); }          //通过指向外围类对象的指针操作hashmap. 如果这个set进行clear操作, 对应的hashmap也会被clear
        public final Iterator<K> iterator()     { return new KeyIterator(); }      //KeySet对象事实上就是在原来的HashMap对象上套了个壳. 迭代器操作什么的都是建立在原来的对象上的
        public final boolean contains(Object o) { return containsKey(o); }         //Values类, EntrySet类同理
        public final boolean remove(Object key) {
            return removeNode(hash(key), key, null, false, true) != null;
        }
        public final Spliterator<K> spliterator() {
            return new KeySpliterator<>(HashMap.this, 0, -1, 0, 0);
        }
        public final void forEach(Consumer<? super K> action) {
            Node<K,V>[] tab;
            if (action == null)
                throw new NullPointerException();
            if (size > 0 && (tab = table) != null) {
                int mc = modCount;
                for (int i = 0; i < tab.length; ++i) {
                    for (Node<K,V> e = tab[i]; e != null; e = e.next)
                        action.accept(e.key);
                }
                if (modCount != mc)
                    throw new ConcurrentModificationException();
            }
        }
    }

    /**
     * Returns a {@link Collection} view of the values contained in this map.
     * The collection is backed by the map, so changes to the map are
     * reflected in the collection, and vice-versa.  If the map is
     * modified while an iteration over the collection is in progress
     * (except through the iterator's own <tt>remove</tt> operation),
     * the results of the iteration are undefined.  The collection
     * supports element removal, which removes the corresponding
     * mapping from the map, via the <tt>Iterator.remove</tt>,
     * <tt>Collection.remove</tt>, <tt>removeAll</tt>,
     * <tt>retainAll</tt> and <tt>clear</tt> operations.  It does not
     * support the <tt>add</tt> or <tt>addAll</tt> operations.
     *
     * @return a view of the values contained in this map
     */
    public Collection<V> values() {
        Collection<V> vs = values;
        if (vs == null) {
            vs = new Values();
            values = vs;
        }
        return vs;
    }

    final class Values extends AbstractCollection<V> {          //同KeySet一样, 返回的是视图而不是副本
        public final int size()                 { return size; }
        public final void clear()               { HashMap.this.clear(); }
        public final Iterator<V> iterator()     { return new ValueIterator(); }
        public final boolean contains(Object o) { return containsValue(o); }
        public final Spliterator<V> spliterator() {
            return new ValueSpliterator<>(HashMap.this, 0, -1, 0, 0);
        }
        public final void forEach(Consumer<? super V> action) {
            Node<K,V>[] tab;
            if (action == null)
                throw new NullPointerException();
            if (size > 0 && (tab = table) != null) {
                int mc = modCount;
                for (int i = 0; i < tab.length; ++i) {
                    for (Node<K,V> e = tab[i]; e != null; e = e.next)
                        action.accept(e.value);
                }
                if (modCount != mc)
                    throw new ConcurrentModificationException();
            }
        }
    }

    /**
     * Returns a {@link Set} view of the mappings contained in this map.
     * The set is backed by the map, so changes to the map are
     * reflected in the set, and vice-versa.  If the map is modified
     * while an iteration over the set is in progress (except through
     * the iterator's own <tt>remove</tt> operation, or through the
     * <tt>setValue</tt> operation on a map entry returned by the
     * iterator) the results of the iteration are undefined.  The set
     * supports element removal, which removes the corresponding
     * mapping from the map, via the <tt>Iterator.remove</tt>,
     * <tt>Set.remove</tt>, <tt>removeAll</tt>, <tt>retainAll</tt> and
     * <tt>clear</tt> operations.  It does not support the
     * <tt>add</tt> or <tt>addAll</tt> operations.
     *
     * @return a set view of the mappings contained in this map
     */
    public Set<Map.Entry<K,V>> entrySet() {
        Set<Map.Entry<K,V>> es;
        return (es = entrySet) == null ? (entrySet = new EntrySet()) : es;
    }

    final class EntrySet extends AbstractSet<Map.Entry<K,V>> {
        public final int size()                 { return size; }
        public final void clear()               { HashMap.this.clear(); }
        public final Iterator<Map.Entry<K,V>> iterator() {
            return new EntryIterator();
        }
        public final boolean contains(Object o) {
            if (!(o instanceof Map.Entry))
                return false;
            Map.Entry<?,?> e = (Map.Entry<?,?>) o;
            Object key = e.getKey();
            Node<K,V> candidate = getNode(hash(key), key);
            return candidate != null && candidate.equals(e);
        }
        public final boolean remove(Object o) {
            if (o instanceof Map.Entry) {
                Map.Entry<?,?> e = (Map.Entry<?,?>) o;
                Object key = e.getKey();
                Object value = e.getValue();
                return removeNode(hash(key), key, value, true, true) != null;
            }
            return false;
        }
        public final Spliterator<Map.Entry<K,V>> spliterator() {
            return new EntrySpliterator<>(HashMap.this, 0, -1, 0, 0);
        }
        public final void forEach(Consumer<? super Map.Entry<K,V>> action) {
            Node<K,V>[] tab;
            if (action == null)
                throw new NullPointerException();
            if (size > 0 && (tab = table) != null) {
                int mc = modCount;
                for (int i = 0; i < tab.length; ++i) {
                    for (Node<K,V> e = tab[i]; e != null; e = e.next)
                        action.accept(e);
                }
                if (modCount != mc)
                    throw new ConcurrentModificationException();
            }
        }
    }

    // Overrides of JDK8 Map extension methods          //这一部分是jdk8新扩展的方法了, 大多数方法和底层原理无关, 就不仔细分析了. 我本人看源码的目的是为了了解底层

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

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

    @Override
    public boolean remove(Object key, Object value) {
        return removeNode(hash(key), key, value, true, true) != null;
    }

    @Override
    public boolean replace(K key, V oldValue, V newValue) {
        Node<K,V> e; V v;
        if ((e = getNode(hash(key), key)) != null &&
            ((v = e.value) == oldValue || (v != null && v.equals(oldValue)))) {
            e.value = newValue;
            afterNodeAccess(e);
            return true;
        }
        return false;
    }

    @Override
    public V replace(K key, V value) {
        Node<K,V> e;
        if ((e = getNode(hash(key), key)) != null) {
            V oldValue = e.value;
            e.value = value;
            afterNodeAccess(e);
            return oldValue;
        }
        return null;
    }

    @Override
    public V computeIfAbsent(K key,
                             Function<? super K, ? extends V> mappingFunction) {
        if (mappingFunction == null)
            throw new NullPointerException();
        int hash = hash(key);
        Node<K,V>[] tab; Node<K,V> first; int n, i;
        int binCount = 0;
        TreeNode<K,V> t = null;
        Node<K,V> old = null;
        if (size > threshold || (tab = table) == null ||
            (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((first = tab[i = (n - 1) & hash]) != null) {
            if (first instanceof TreeNode)
                old = (t = (TreeNode<K,V>)first).getTreeNode(hash, key);
            else {
                Node<K,V> e = first; K k;
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k)))) {
                        old = e;
                        break;
                    }
                    ++binCount;
                } while ((e = e.next) != null);
            }
            V oldValue;
            if (old != null && (oldValue = old.value) != null) {
                afterNodeAccess(old);
                return oldValue;
            }
        }
        V v = mappingFunction.apply(key);
        if (v == null) {
            return null;
        } else if (old != null) {
            old.value = v;
            afterNodeAccess(old);
            return v;
        }
        else if (t != null)
            t.putTreeVal(this, tab, hash, key, v);
        else {
            tab[i] = newNode(hash, key, v, first);
            if (binCount >= TREEIFY_THRESHOLD - 1)
                treeifyBin(tab, hash);
        }
        ++modCount;
        ++size;
        afterNodeInsertion(true);
        return v;
    }

    public V computeIfPresent(K key,
                              BiFunction<? super K, ? super V, ? extends V> remappingFunction) {
        if (remappingFunction == null)
            throw new NullPointerException();
        Node<K,V> e; V oldValue;
        int hash = hash(key);
        if ((e = getNode(hash, key)) != null &&
            (oldValue = e.value) != null) {
            V v = remappingFunction.apply(key, oldValue);
            if (v != null) {
                e.value = v;
                afterNodeAccess(e);
                return v;
            }
            else
                removeNode(hash, key, null, false, true);
        }
        return null;
    }

    @Override
    public V compute(K key,
                     BiFunction<? super K, ? super V, ? extends V> remappingFunction) {
        if (remappingFunction == null)
            throw new NullPointerException();
        int hash = hash(key);
        Node<K,V>[] tab; Node<K,V> first; int n, i;
        int binCount = 0;
        TreeNode<K,V> t = null;
        Node<K,V> old = null;
        if (size > threshold || (tab = table) == null ||
            (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((first = tab[i = (n - 1) & hash]) != null) {
            if (first instanceof TreeNode)
                old = (t = (TreeNode<K,V>)first).getTreeNode(hash, key);
            else {
                Node<K,V> e = first; K k;
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k)))) {
                        old = e;
                        break;
                    }
                    ++binCount;
                } while ((e = e.next) != null);
            }
        }
        V oldValue = (old == null) ? null : old.value;
        V v = remappingFunction.apply(key, oldValue);
        if (old != null) {
            if (v != null) {
                old.value = v;
                afterNodeAccess(old);
            }
            else
                removeNode(hash, key, null, false, true);
        }
        else if (v != null) {
            if (t != null)
                t.putTreeVal(this, tab, hash, key, v);
            else {
                tab[i] = newNode(hash, key, v, first);
                if (binCount >= TREEIFY_THRESHOLD - 1)
                    treeifyBin(tab, hash);
            }
            ++modCount;
            ++size;
            afterNodeInsertion(true);
        }
        return v;
    }

    @Override
    public V merge(K key, V value,
                   BiFunction<? super V, ? super V, ? extends V> remappingFunction) {
        if (value == null)
            throw new NullPointerException();
        if (remappingFunction == null)
            throw new NullPointerException();
        int hash = hash(key);
        Node<K,V>[] tab; Node<K,V> first; int n, i;
        int binCount = 0;
        TreeNode<K,V> t = null;
        Node<K,V> old = null;
        if (size > threshold || (tab = table) == null ||
            (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((first = tab[i = (n - 1) & hash]) != null) {
            if (first instanceof TreeNode)
                old = (t = (TreeNode<K,V>)first).getTreeNode(hash, key);
            else {
                Node<K,V> e = first; K k;
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k)))) {
                        old = e;
                        break;
                    }
                    ++binCount;
                } while ((e = e.next) != null);
            }
        }
        if (old != null) {
            V v;
            if (old.value != null)
                v = remappingFunction.apply(old.value, value);
            else
                v = value;
            if (v != null) {
                old.value = v;
                afterNodeAccess(old);
            }
            else
                removeNode(hash, key, null, false, true);
            return v;
        }
        if (value != null) {
            if (t != null)
                t.putTreeVal(this, tab, hash, key, value);
            else {
                tab[i] = newNode(hash, key, value, first);
                if (binCount >= TREEIFY_THRESHOLD - 1)
                    treeifyBin(tab, hash);
            }
            ++modCount;
            ++size;
            afterNodeInsertion(true);
        }
        return value;
    }

    @Override
    public void forEach(BiConsumer<? super K, ? super V> action) {
        Node<K,V>[] tab;
        if (action == null)
            throw new NullPointerException();
        if (size > 0 && (tab = table) != null) {
            int mc = modCount;
            for (int i = 0; i < tab.length; ++i) {
                for (Node<K,V> e = tab[i]; e != null; e = e.next)
                    action.accept(e.key, e.value);
            }
            if (modCount != mc)
                throw new ConcurrentModificationException();
        }
    }

    @Override
    public void replaceAll(BiFunction<? super K, ? super V, ? extends V> function) {
        Node<K,V>[] tab;
        if (function == null)
            throw new NullPointerException();
        if (size > 0 && (tab = table) != null) {
            int mc = modCount;
            for (int i = 0; i < tab.length; ++i) {
                for (Node<K,V> e = tab[i]; e != null; e = e.next) {
                    e.value = function.apply(e.key, e.value);
                }
            }
            if (modCount != mc)
                throw new ConcurrentModificationException();
        }
    }

    /* ------------------------------------------------------------ */
    // Cloning and serialization

    /**
     * Returns a shallow copy of this <tt>HashMap</tt> instance: the keys and
     * values themselves are not cloned.
     *
     * @return a shallow copy of this map
     */
    @SuppressWarnings("unchecked")
    @Override
    public Object clone() {
        HashMap<K,V> result;
        try {
            result = (HashMap<K,V>)super.clone();
        } catch (CloneNotSupportedException e) {
            // this shouldn't happen, since we are Cloneable
            throw new InternalError(e);
        }
        result.reinitialize();
        result.putMapEntries(this, false);
        return result;
    }

    // These methods are also used when serializing HashSets
    final float loadFactor() { return loadFactor; }
    final int capacity() {
        return (table != null) ? table.length :
            (threshold > 0) ? threshold :
            DEFAULT_INITIAL_CAPACITY;
    }

    /**
     * Save the state of the <tt>HashMap</tt> instance to a stream (i.e.,
     * serialize it).
     *
     * @serialData The <i>capacity</i> of the HashMap (the length of the
     *             bucket array) is emitted (int), followed by the
     *             <i>size</i> (an int, the number of key-value
     *             mappings), followed by the key (Object) and value (Object)
     *             for each key-value mapping.  The key-value mappings are
     *             emitted in no particular order.
     */
    private void writeObject(java.io.ObjectOutputStream s)                   //写入的时候只写入桶数量, size, 和各个entry. 其他的不写
        throws IOException {
        int buckets = capacity();
        // Write out the threshold, loadfactor, and any hidden stuff
        s.defaultWriteObject();
        s.writeInt(buckets);
        s.writeInt(size);
        internalWriteEntries(s);
    }
//之前一直好奇writeObject和readObject方法是怎么回事, 是private的并且没有public方法调用他们. 如果写这两个方法是为了ObjectInputStream或ObjectOutputStream使用, 那也不对啊.
//这两个流使用序列化的时候是ObjectInputStream类中的方法 ObjectInputStream in = new ObjectInputStream(); in.readObject();
// 而不是HashMap类中的方法HashMap map = new HashMap(); map.readObject(in);
//后来上网查了资料 https://zhuanlan.zhihu.com/p/84533476 才发现ObjectInputStream类中的readObject方法也是通过反射看看被写入的类中有没有实现.
//实际上在ObjectOutputStream中进行序列化操作的时候，会判断被序列化的对象是否自己重写了writeObject方法，如果重写了，就会调用被序列化对象自己的writeObject方法，如果没有重写，才会调用默认的序列化方法。
    /**
     * Reconstitutes this map from a stream (that is, deserializes it).
     * @param s the stream
     * @throws ClassNotFoundException if the class of a serialized object
     *         could not be found
     * @throws IOException if an I/O error occurs
     */
    private void readObject(java.io.ObjectInputStream s)              //读取的时候只读键值对数量, 和各个键值对
        throws IOException, ClassNotFoundException {
        // Read in the threshold (ignored), loadfactor, and any hidden stuff
        s.defaultReadObject();
        reinitialize();
        if (loadFactor <= 0 || Float.isNaN(loadFactor))
            throw new InvalidObjectException("Illegal load factor: " +
                                             loadFactor);
        s.readInt();                // Read and ignore number of buckets
        int mappings = s.readInt(); // Read number of mappings (size)
        if (mappings < 0)
            throw new InvalidObjectException("Illegal mappings count: " +
                                             mappings);
        else if (mappings > 0) { // (if zero, use defaults)
            // Size the table using given load factor only if within
            // range of 0.25...4.0
            float lf = Math.min(Math.max(0.25f, loadFactor), 4.0f);
            float fc = (float)mappings / lf + 1.0f;
            int cap = ((fc < DEFAULT_INITIAL_CAPACITY) ?
                       DEFAULT_INITIAL_CAPACITY :
                       (fc >= MAXIMUM_CAPACITY) ?
                       MAXIMUM_CAPACITY :
                       tableSizeFor((int)fc));
            float ft = (float)cap * lf;
            threshold = ((cap < MAXIMUM_CAPACITY && ft < MAXIMUM_CAPACITY) ?
                         (int)ft : Integer.MAX_VALUE);

            // Check Map.Entry[].class since it's the nearest public type to
            // what we're actually creating.
            SharedSecrets.getJavaOISAccess().checkArray(s, Map.Entry[].class, cap);
            @SuppressWarnings({"rawtypes","unchecked"})
            Node<K,V>[] tab = (Node<K,V>[])new Node[cap];
            table = tab;

            // Read the keys and values, and put the mappings in the HashMap
            for (int i = 0; i < mappings; i++) {
                @SuppressWarnings("unchecked")
                    K key = (K) s.readObject();
                @SuppressWarnings("unchecked")
                    V value = (V) s.readObject();
                putVal(hash(key), key, value, false, false);
            }
        }
    }

    /* ------------------------------------------------------------ */
    // iterators

    abstract class HashIterator {                               //entrySet(), keySet(), values()返回的集合的迭代器的基类. 从第0个bucket开始遍历
        Node<K,V> next;        // next entry to return
        Node<K,V> current;     // current entry
        int expectedModCount;  // for fast-fail
        int index;             // current slot

        HashIterator() {
            expectedModCount = modCount;
            Node<K,V>[] t = table;
            current = next = null;
            index = 0;
            if (t != null && size > 0) { // advance to first entry
                do {} while (index < t.length && (next = t[index++]) == null);
            }
        }

        public final boolean hasNext() {
            return next != null;
        }

        final Node<K,V> nextNode() {
            Node<K,V>[] t;
            Node<K,V> e = next;
            if (modCount != expectedModCount)
                throw new ConcurrentModificationException();
            if (e == null)
                throw new NoSuchElementException();
            if ((next = (current = e).next) == null && (t = table) != null) {        //把当前的next传递给current, 然后找下一个next
                do {} while (index < t.length && (next = t[index++]) == null);
            }
            return e;
        }

        public final void remove() {
            Node<K,V> p = current;
            if (p == null)
                throw new IllegalStateException();
            if (modCount != expectedModCount)
                throw new ConcurrentModificationException();
            current = null;
            K key = p.key;
            removeNode(hash(key), key, null, false, false);
            expectedModCount = modCount;
        }
    }

    final class KeyIterator extends HashIterator
        implements Iterator<K> {
        public final K next() { return nextNode().key; }
    }

    final class ValueIterator extends HashIterator
        implements Iterator<V> {
        public final V next() { return nextNode().value; }
    }

    final class EntryIterator extends HashIterator
        implements Iterator<Map.Entry<K,V>> {
        public final Map.Entry<K,V> next() { return nextNode(); }
    }

    /* ------------------------------------------------------------ */
    // spliterators                                  //为并行遍历设置的迭代器

    static class HashMapSpliterator<K,V> {
        final HashMap<K,V> map;
        Node<K,V> current;          // current node
        int index;                  // current index, modified on advance/split
        int fence;                  // one past last index
        int est;                    // size estimate
        int expectedModCount;       // for comodification checks

        HashMapSpliterator(HashMap<K,V> m, int origin,
                           int fence, int est,
                           int expectedModCount) {
            this.map = m;
            this.index = origin;
            this.fence = fence;
            this.est = est;
            this.expectedModCount = expectedModCount;
        }

        final int getFence() { // initialize fence and size on first use
            int hi;
            if ((hi = fence) < 0) {
                HashMap<K,V> m = map;
                est = m.size;
                expectedModCount = m.modCount;
                Node<K,V>[] tab = m.table;
                hi = fence = (tab == null) ? 0 : tab.length;
            }
            return hi;
        }

        public final long estimateSize() {
            getFence(); // force init
            return (long) est;
        }
    }

    static final class KeySpliterator<K,V>
        extends HashMapSpliterator<K,V>
        implements Spliterator<K> {
        KeySpliterator(HashMap<K,V> m, int origin, int fence, int est,
                       int expectedModCount) {
            super(m, origin, fence, est, expectedModCount);
        }

        public KeySpliterator<K,V> trySplit() {
            int hi = getFence(), lo = index, mid = (lo + hi) >>> 1;
            return (lo >= mid || current != null) ? null :
                new KeySpliterator<>(map, lo, index = mid, est >>>= 1,
                                        expectedModCount);
        }

        public void forEachRemaining(Consumer<? super K> action) {
            int i, hi, mc;
            if (action == null)
                throw new NullPointerException();
            HashMap<K,V> m = map;
            Node<K,V>[] tab = m.table;
            if ((hi = fence) < 0) {
                mc = expectedModCount = m.modCount;
                hi = fence = (tab == null) ? 0 : tab.length;
            }
            else
                mc = expectedModCount;
            if (tab != null && tab.length >= hi &&
                (i = index) >= 0 && (i < (index = hi) || current != null)) {
                Node<K,V> p = current;
                current = null;
                do {
                    if (p == null)
                        p = tab[i++];
                    else {
                        action.accept(p.key);
                        p = p.next;
                    }
                } while (p != null || i < hi);
                if (m.modCount != mc)
                    throw new ConcurrentModificationException();
            }
        }

        public boolean tryAdvance(Consumer<? super K> action) {
            int hi;
            if (action == null)
                throw new NullPointerException();
            Node<K,V>[] tab = map.table;
            if (tab != null && tab.length >= (hi = getFence()) && index >= 0) {
                while (current != null || index < hi) {
                    if (current == null)
                        current = tab[index++];
                    else {
                        K k = current.key;
                        current = current.next;
                        action.accept(k);
                        if (map.modCount != expectedModCount)
                            throw new ConcurrentModificationException();
                        return true;
                    }
                }
            }
            return false;
        }

        public int characteristics() {
            return (fence < 0 || est == map.size ? Spliterator.SIZED : 0) |
                Spliterator.DISTINCT;
        }
    }

    static final class ValueSpliterator<K,V>
        extends HashMapSpliterator<K,V>
        implements Spliterator<V> {
        ValueSpliterator(HashMap<K,V> m, int origin, int fence, int est,
                         int expectedModCount) {
            super(m, origin, fence, est, expectedModCount);
        }

        public ValueSpliterator<K,V> trySplit() {
            int hi = getFence(), lo = index, mid = (lo + hi) >>> 1;
            return (lo >= mid || current != null) ? null :
                new ValueSpliterator<>(map, lo, index = mid, est >>>= 1,
                                          expectedModCount);
        }

        public void forEachRemaining(Consumer<? super V> action) {
            int i, hi, mc;
            if (action == null)
                throw new NullPointerException();
            HashMap<K,V> m = map;
            Node<K,V>[] tab = m.table;
            if ((hi = fence) < 0) {
                mc = expectedModCount = m.modCount;
                hi = fence = (tab == null) ? 0 : tab.length;
            }
            else
                mc = expectedModCount;
            if (tab != null && tab.length >= hi &&
                (i = index) >= 0 && (i < (index = hi) || current != null)) {
                Node<K,V> p = current;
                current = null;
                do {
                    if (p == null)
                        p = tab[i++];
                    else {
                        action.accept(p.value);
                        p = p.next;
                    }
                } while (p != null || i < hi);
                if (m.modCount != mc)
                    throw new ConcurrentModificationException();
            }
        }

        public boolean tryAdvance(Consumer<? super V> action) {
            int hi;
            if (action == null)
                throw new NullPointerException();
            Node<K,V>[] tab = map.table;
            if (tab != null && tab.length >= (hi = getFence()) && index >= 0) {
                while (current != null || index < hi) {
                    if (current == null)
                        current = tab[index++];
                    else {
                        V v = current.value;
                        current = current.next;
                        action.accept(v);
                        if (map.modCount != expectedModCount)
                            throw new ConcurrentModificationException();
                        return true;
                    }
                }
            }
            return false;
        }

        public int characteristics() {
            return (fence < 0 || est == map.size ? Spliterator.SIZED : 0);
        }
    }

    static final class EntrySpliterator<K,V>
        extends HashMapSpliterator<K,V>
        implements Spliterator<Map.Entry<K,V>> {
        EntrySpliterator(HashMap<K,V> m, int origin, int fence, int est,
                         int expectedModCount) {
            super(m, origin, fence, est, expectedModCount);
        }

        public EntrySpliterator<K,V> trySplit() {
            int hi = getFence(), lo = index, mid = (lo + hi) >>> 1;
            return (lo >= mid || current != null) ? null :
                new EntrySpliterator<>(map, lo, index = mid, est >>>= 1,
                                          expectedModCount);
        }

        public void forEachRemaining(Consumer<? super Map.Entry<K,V>> action) {
            int i, hi, mc;
            if (action == null)
                throw new NullPointerException();
            HashMap<K,V> m = map;
            Node<K,V>[] tab = m.table;
            if ((hi = fence) < 0) {
                mc = expectedModCount = m.modCount;
                hi = fence = (tab == null) ? 0 : tab.length;
            }
            else
                mc = expectedModCount;
            if (tab != null && tab.length >= hi &&
                (i = index) >= 0 && (i < (index = hi) || current != null)) {
                Node<K,V> p = current;
                current = null;
                do {
                    if (p == null)
                        p = tab[i++];
                    else {
                        action.accept(p);
                        p = p.next;
                    }
                } while (p != null || i < hi);
                if (m.modCount != mc)
                    throw new ConcurrentModificationException();
            }
        }

        public boolean tryAdvance(Consumer<? super Map.Entry<K,V>> action) {
            int hi;
            if (action == null)
                throw new NullPointerException();
            Node<K,V>[] tab = map.table;
            if (tab != null && tab.length >= (hi = getFence()) && index >= 0) {
                while (current != null || index < hi) {
                    if (current == null)
                        current = tab[index++];
                    else {
                        Node<K,V> e = current;
                        current = current.next;
                        action.accept(e);
                        if (map.modCount != expectedModCount)
                            throw new ConcurrentModificationException();
                        return true;
                    }
                }
            }
            return false;
        }

        public int characteristics() {
            return (fence < 0 || est == map.size ? Spliterator.SIZED : 0) |
                Spliterator.DISTINCT;
        }
    }

    /* ------------------------------------------------------------ */
    // LinkedHashMap support


    /*
     * The following package-protected methods are designed to be
     * overridden by LinkedHashMap, but not by any other subclass.
     * Nearly all other internal methods are also package-protected
     * but are declared final, so can be used by LinkedHashMap, view
     * classes, and HashSet.
     */

    // Create a regular (non-tree) node
    Node<K,V> newNode(int hash, K key, V value, Node<K,V> next) {
        return new Node<>(hash, key, value, next);
    }

    // For conversion from TreeNodes to plain nodes
    Node<K,V> replacementNode(Node<K,V> p, Node<K,V> next) {
        return new Node<>(p.hash, p.key, p.value, next);
    }

    // Create a tree bin node
    TreeNode<K,V> newTreeNode(int hash, K key, V value, Node<K,V> next) {
        return new TreeNode<>(hash, key, value, next);
    }

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

    /**
     * Reset to initial default state.  Called by clone and readObject.
     */
    void reinitialize() {
        table = null;
        entrySet = null;
        keySet = null;
        values = null;
        modCount = 0;
        threshold = 0;
        size = 0;
    }

    // Callbacks to allow LinkedHashMap post-actions
    void afterNodeAccess(Node<K,V> p) { }
    void afterNodeInsertion(boolean evict) { }
    void afterNodeRemoval(Node<K,V> p) { }

    // Called only from writeObject, to ensure compatible ordering.
    void internalWriteEntries(java.io.ObjectOutputStream s) throws IOException {
        Node<K,V>[] tab;
        if (size > 0 && (tab = table) != null) {
            for (int i = 0; i < tab.length; ++i) {
                for (Node<K,V> e = tab[i]; e != null; e = e.next) {
                    s.writeObject(e.key);
                    s.writeObject(e.value);
                }
            }
        }
    }

    /* ------------------------------------------------------------ */
    // Tree bins
    /**
     *  LinkedHashMap.Entry的定义
    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);
        }
    }
     */

    /**
     * Entry for Tree bins. Extends LinkedHashMap.Entry (which in turn
     * extends Node) so can be used as extension of either regular or
     * linked node.
     */     //这里推荐 https://blog.csdn.net/anlian523/article/details/103649200 这篇博客, 对红黑树的操作讲解的都非常细. 我的笔记有一部分也是参考的这篇博客
    static final class TreeNode<K,V> extends LinkedHashMap.Entry<K,V> {                 // 树状的bin节点
        TreeNode<K,V> parent;  // red-black tree links     //是这样的继承关系, LinkedHashMap.Entry继承了HashMap.Node, HashMap.TreeNode又继承了LinkedHashMap.Entry
        TreeNode<K,V> left;
        TreeNode<K,V> right;
        TreeNode<K,V> prev;    // needed to unlink next upon deletion        链表的上一个节点
        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;
            }
        }

        /**
         * Ensures that the given root is the first node of its bin.
         */ //因为有的时候旋转会使根节点发生变化, 即经过某次操作后, table[]中储存的可能不是根节点了. 这时要重新将根节点放到table[]中
        static <K,V> void moveRootToFront(Node<K,V>[] tab, TreeNode<K,V> root) {
            int n;
            if (root != null && tab != null && (n = tab.length) > 0) {
                int index = (n - 1) & root.hash;
                TreeNode<K,V> first = (TreeNode<K,V>)tab[index];
                if (root != first) {
                    Node<K,V> rn; 
                    tab[index] = root;                                              //第一步, 先把table[index]设为root
                    TreeNode<K,V> rp = root.prev;//rn代表root.prev, rn代表root.next
                    if ((rn = root.next) != null)                                  //第二步, 处理链表顺序使得root在双向链表的第一个
                        ((TreeNode<K,V>)rn).prev = rp;                             // 即链表 first <-> ... <-> rp <-> root <-> rn <-> ...
                    if (rp != null)                                                //变为 root <-> first <-> ... <-> rp <-> rn <-> ...
                        rp.next = rn; 
                    if (first != null)
                        first.prev = root;
                    root.next = first;
                    root.prev = null;
                }
                assert checkInvariants(root);
            }
        }

        /**
         * Finds the node starting at root p with the given hash and key.
         * The kc argument caches comparableClassFor(key) upon first use
         * comparing keys.
         */ 
        //这段代码大概的逻辑如下
        //  TreeNode find(Object k)
        //  {
        //      TreeNode curr = this;
        //      while(curr != null)
        //      {
        //         if(curr.key.hash > k.hash)
        //             curr = curr.left;             //红黑树的顺序貌似是先以hashcode为主, hashcode相等再去调用对象的compareTo方法
        //         else if(curr.key.hash < k.hash)
        //             curr = curr.left;
        //         else if(curr.key.equals(k))
        //             return curr;
        //         else if(k有compareTo方法 && k.compareTo(curr.key) != 0)                         //hashcode相等, 但是值不相等, 所以要用compareTo方法比较curr和 k
        //         {                                    //正常情况下不可能出现 k.compareTo(curr.key) == 0的. 因为这两个如果相等的话, 在上一层的if中就会被返回.
        //             if(curr.key.compareTo(k) > 0)
        //                 curr = curr.left;
        //             else
        //                 curr = curr.right;
        //         }
        //         else if(curr.right.find(k) != null)                 //hashCode和compareTo都判断不出来大小关系, 只能先递归地判断右边整个子树有没有这个元素, 然后再迭代的查找左子树
        //             return curr.right.find(k);
        //         else      //右子树也没有, 查找左子树
        //             curr = curr.left;
        //      }
        //      return null;
        //  }
        
        final TreeNode<K,V> find(int h, Object k, Class<?> kc) {         //kc是对象k的Class对象
            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)  //如果k没有实现Comparable接口或者compareTo方法返回的是0, 就递归的查找右子树的所有值
                    p = (dir < 0) ? pl : pr;
                else if ((q = pr.find(h, k, kc)) != null)
                    return q;
                else
                    p = pl;                                          // 如果右子树也没有, 就查找左子树.
            } while (p != null);                                     
            return null;
        }

        /**
         * Calls find for root node.
         */
        final TreeNode<K,V> getTreeNode(int h, Object k) {
            return ((parent != null) ? root() : this).find(h, k, null);
        }

        /**
         * Tie-breaking utility for ordering insertions when equal
         * hashCodes and non-comparable. We don't require a total
         * order, just a consistent insertion rule to maintain
         * equivalence across rebalancings. Tie-breaking further than
         * necessary simplifies testing a bit.                   //这个函数只有两个对象hash相同, compareTo相同(或者不支持compareTo方法)时在才会调用
         */ 
        static int tieBreakOrder(Object a, Object b) {   //可以看成一个约定, 当两个对象hash相同, compareTo相同时, 通过内存中的位置规定了插入红黑树的方向. 是左孩子还是右孩子
            int d;                                   //但是查找的时候不能这么做, 因为查找的时候可能是不同地址的相同对象, 如果插入时的对象和查找时给的对象相等, 但地址值不一样, 就会发生错误
            if (a == null || b == null ||            //查找的时候如果发生了hash相同, compareTo相同(或者不支持compareTo方法)的情况, 只能左右子树一起查, 不能根据这个tieBreakOrder方法只查一个子树
                (d = a.getClass().getName().
                 compareTo(b.getClass().getName())) == 0)
                d = (System.identityHashCode(a) <= System.identityHashCode(b) ?
                     -1 : 1);
            return d;
        }

        /**
         * Forms tree of the nodes linked from this node.
         */
        final void treeify(Node<K,V>[] tab) {                        //这个方法虽然长, 但是完成的事情很简单, 只是在红黑树中找到x合适的位置(叶子节点)并插入
            TreeNode<K,V> root = null;                               //至于插入后怎么旋转, 怎么改变红黑颜色, 就是balanceInsertion()方法的事情了
            for (TreeNode<K,V> x = this, next; x != null; x = next) {
                next = (TreeNode<K,V>)x.next;
                x.left = x.right = null;
                if (root == null) {                               //第一个节点作为头节点
                    x.parent = null;
                    x.red = false;
                    root = x;
                }
                else {
                    K k = x.key;
                    int h = x.hash;
                    Class<?> kc = null;
                    for (TreeNode<K,V> p = root;;) {
                        int dir, ph;
                        K pk = p.key;
                        if ((ph = p.hash) > h)
                            dir = -1;                       // 又是熟悉的套路, 先根据hash确定在左子树还是右子树, hash相同在判断compareTo方法
                        else if (ph < h)                    //compareTo相同或者没有compareTo方法, 就按照我们插入时的约定tieBreakOrder, 根据内存地址来判断
                            dir = 1;                        //如果a的内存地址小于等于b的内存地址, 就把a放到b的左子树中的某个叶子节点的位置. 这里的dir是方向direction的意思. 即走左子树还是右子树
                        else if ((kc == null &&
                                  (kc = comparableClassFor(k)) == null) ||
                                 (dir = compareComparables(kc, k, pk)) == 0)
                            dir = tieBreakOrder(k, pk);

                        TreeNode<K,V> xp = p;
                        if ((p = (dir <= 0) ? p.left : p.right) == null) {             //注意到这个==null, 表示只能在叶子节点上插入
                            x.parent = xp;
                            if (dir <= 0)
                                xp.left = x;
                            else
                                xp.right = x;
                            root = balanceInsertion(root, x);
                            break;
                        }
                    }
                }
            }
            moveRootToFront(tab, root);
        }

        /**
         * Returns a list of non-TreeNodes replacing those linked from
         * this node.
         */
        final Node<K,V> untreeify(HashMap<K,V> map) {          //这个方法就很简单了
            Node<K,V> hd = null, tl = null;
            for (Node<K,V> q = this; q != null; q = q.next) {
                Node<K,V> p = map.replacementNode(q, null);
                if (tl == null)
                    hd = p;
                else
                    tl.next = p;
                tl = p;
            }
            return hd;
        }

        /**
         * Tree version of putVal.
         */
        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;;) {                          //和treeify方法差不多, 都是先判断hash值, 再用compareTo, 这两个都判断不出来的时候, 就要左右两棵子树都查找了
                int dir, ph; K pk;                                   //一旦进行左子树和右子树的全树扫描, 就把search标记为true. 
                if ((ph = p.hash) > h)                               //下一次再出现这样的情况, 就说明树里面没有k这个键, 按照tieBreakOrder的规则找个地方插入就可以了
                    dir = -1;
                else if (ph < h)
                    dir = 1;
                else if ((pk = p.key) == k || (k != null && k.equals(pk)))
                    return p;                      //这里直接返回键k对应的TreeNode, 由调用这个方法的调用者把p的value改成v.
                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) {      //如果搜索到红黑树的叶子节点还是没有找到, 就在叶子节点插入新的TreeNode x.
                    Node<K,V> xpn = xp.next;                            //至于插入后红黑树的旋转和重新染色, 要调用静态方法balanceInsertion()了
                    TreeNode<K,V> x = map.newTreeNode(h, k, v, xpn);    //同时, 要把x加入到双向链表中, 放到x.parent的后面
                    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;
                }
            }
        }

        /**
         * Removes the given node, that must be present before this call.
         * This is messier than typical red-black deletion code because we
         * cannot swap the contents of an interior node with a leaf
         * successor that is pinned by "next" pointers that are accessible       //由于TreeNode除了left, 和right指针外, 还有prev和next指针. 所以不能像普通的红黑树一样删除节点
         * independently during traversal. So instead we swap the tree
         * linkages. If the current tree appears to have too few nodes,
         * the bin is converted back to a plain bin. (The test triggers
         * somewhere between 2 and 6 nodes, depending on tree structure).
         */
        final void removeTreeNode(HashMap<K,V> map, Node<K,V>[] tab,          //从tree中移除当前的树节点this.
                                  boolean movable) {
            int n;
            if (tab == null || (n = tab.length) == 0)            
                return;
            int index = (n - 1) & hash;          
            TreeNode<K,V> first = (TreeNode<K,V>)tab[index], root = first, rl;  //first为桶中的第一个元素
            TreeNode<K,V> succ = (TreeNode<K,V>)next, pred = prev;             //这里的succ为当前节点的successor, pred为当前节点的predecessor
            if (pred == null)
                tab[index] = first = succ;
            else
                pred.next = succ;
            if (succ != null)
                succ.prev = pred;
            if (first == null)          //如果删除了this之后整个树为空了就直接返回
                return;
            if (root.parent != null)   //重置根节点
                root = root.root();
            if (root == null
                || (movable
                    && (root.right == null         //如果右子树是空, 左子树不会超过1个元素(因为是红黑树). 所以要非树化
                        || (rl = root.left) == null  //如果左子树是空, 同样要非树化
                        || rl.left == null))) {       //如果左子树的左子树是空, 左子树的右子树不会超过1个元素, 所以左子树不会超过2个元素, 右子树自然不可能超过4个元素
                tab[index] = first.untreeify(map);  // too small
                return;
            }
            TreeNode<K,V> p = this, pl = left, pr = right, replacement;          //p为要删除的节点
            if (pl != null && pr != null) {                                     //第一种情况  p的左右子树都不为空
                TreeNode<K,V> s = pr, sl;                                        //从右子树中找到右子树最小的节点s来替代即将要被删除的节点p
                while ((sl = s.left) != null) // find successor
                    s = sl;
                boolean c = s.red; s.red = p.red; p.red = c; // swap colors      //交换s和p的颜色
                TreeNode<K,V> sr = s.right;                          //提前备份s的右节点, 和p的父节点
                TreeNode<K,V> pp = p.parent;
                if (s == pr) { // p was s's direct parent                //如果s就是p的右孩子, 把s设置为p的父节点
                    p.parent = s;                               
                    s.right = p;
                }                            // 这一段是比较难想象的, 画个图会清楚很多. 博客 https://blog.csdn.net/anlian523/article/details/103649200 上面写的十分好 
                else {
                    TreeNode<K,V> sp = s.parent;
                    if ((p.parent = sp) != null) {        //p放到s的位置上, 这一段这么多的交换操作, 实际上就是为了交换s和p的位置
                        if (s == sp.left)
                            sp.left = p;                            //        pp                                    pp
                        else                                       //        /                                     /
                            sp.right = p;                         //        p                                     s
                    }                                            //        / \                                   / \
                    if ((s.right = pr) != null)                 //        pl  pr             变为               pl  pr
                        pr.parent = s;                         //            /                                     /
                }                                             //            s                                     p
                p.left = null;                               //              \                                     \
                if ((p.right = sr) != null)                 //                sr                                    sr
                    sr.parent = p;
                if ((s.left = pl) != null)                           //然后将sr赋值给replacement
                    pl.parent = s;
                if ((s.parent = pp) == null)
                    root = s;
                else if (p == pp.left)
                    pp.left = s;
                else
                    pp.right = s;
                if (sr != null)
                    replacement = sr;
                else
                    replacement = p;
            }
            else if (pl != null)                         //第二种情况, p的右子树空, replacement = pl
                replacement = pl;
            else if (pr != null)                        //第三种情况, p的左子树空, replacement = pr
                replacement = pr;
            else                                         //第四种情况, p是叶子节点, replacement = p
                replacement = p;
            if (replacement != p) {                      //将p从树中分离出来, 使得树中没有任何一个节点的指针指向p. (但是p的指针可能还指向树中的某些节点)
                TreeNode<K,V> pp = replacement.parent = p.parent;    //replacement.parent一开始是指向p的, 现在指向p的父节点了
                if (pp == null)
                    root = replacement;
                else if (p == pp.left)
                    pp.left = replacement;                         //pp的孩子也由p变为replacement了. 这样p就分离出来了
                else
                    pp.right = replacement;
                p.left = p.right = p.parent = null;
            }
                                                                                        //p是红色就可以直接删, 不会影响红黑树结构
            TreeNode<K,V> r = p.red ? root : balanceDeletion(root, replacement);        //如果p为黑, 删掉p之后会引起红黑树不满足规则, 从root到任意一个叶子节点的黑色节点数量不变
                                                                                        //, 所以要进行调整, 而replacement就是要被调整的节点
            if (replacement == p) {  // detach            //断绝p和树中任何节点的关系
                TreeNode<K,V> pp = p.parent;
                p.parent = null;
                if (pp != null) {
                    if (p == pp.left)
                        pp.left = null;
                    else if (p == pp.right)
                        pp.right = null;
                }
            }
            if (movable)
                moveRootToFront(tab, r);
        }

        /**
         * 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
         */
        final void split(HashMap<K,V> map, Node<K,V>[] tab, int index, int bit) {         //和resize()方法中的思想差不多, 都是讲一棵树中的元素分为2组, 装到不同的桶中
            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;
                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;
                }
            }

            if (loHead != null) {
                if (lc <= UNTREEIFY_THRESHOLD)
                    tab[index] = loHead.untreeify(map);
                else {
                    tab[index] = loHead;
                    if (hiHead != null) // (else is already treeified)  // 如果high里也有节点，说明low和high二者的红黑树结构由于拆分都应该被破坏掉了，所以需要树化
                    //反之，如果high里没有节点，那说明所有节点都在low里面。又由于连接链表时是按照原顺序来的，而原顺序又保持了
                    //红黑树结构（前提节点都在一个桶里）。所以就不需要做树化操作了，因为红黑树结构没有破坏。
                        loHead.treeify(tab);
                }
            }
            if (hiHead != null) {
                if (hc <= UNTREEIFY_THRESHOLD)
                    tab[index + bit] = hiHead.untreeify(map);
                else {
                    tab[index + bit] = hiHead;
                    if (loHead != null)
                        hiHead.treeify(tab);
                }
            }
        }

        /* ------------------------------------------------------------ */
        // Red-black tree methods, all adapted from CLR

        static <K,V> TreeNode<K,V> rotateLeft(TreeNode<K,V> root,
                                              TreeNode<K,V> p) {         //左旋
            TreeNode<K,V> r, pp, rl;                                    //     p                          pr
            if (p != null && (r = p.right) != null) {                  //    /  \                        / \
                if ((rl = p.right = r.left) != null)                  //    pl  pr            变为      p  rr       如果p.parent不存在, 说明新的根为pr, 肯定是黑色
                    rl.parent = p;                                   //         / \                   / \           
                if ((pp = r.parent = p.parent) == null)             //         rl rr                 pl rl
                    (root = r).red = false;
                else if (pp.left == p)
                    pp.left = r;
                else
                    pp.right = r;
                r.left = p;
                p.parent = r;
            }
            return root;
        }

        static <K,V> TreeNode<K,V> rotateRight(TreeNode<K,V> root,
                                               TreeNode<K,V> p) {
            TreeNode<K,V> l, pp, lr;
            if (p != null && (l = p.left) != null) {               //右旋
                if ((lr = p.left = l.right) != null)              //     p                          pl
                    lr.parent = p;                               //    /  \                        / \
                if ((pp = l.parent = p.parent) == null)         //    pl  pr            变为      ll  p       如果p.parent不存在, 说明新的根为pl, 肯定是黑色
                    (root = l).red = false;                    //    / \                             / \           
                else if (pp.right == p)                       //   ll  lr                          lr  pr
                    pp.right = l;
                else
                    pp.left = l;
                l.right = p;
                p.parent = l;
            }
            return root;
        }

        static <K,V> TreeNode<K,V> balanceInsertion(TreeNode<K,V> root,       //这个函数是修复插入节点之后的红黑树的, 而不是插入节点
                                                    TreeNode<K,V> x) {
            x.red = true;                                                  //新插入的节点总是红的
            for (TreeNode<K,V> xp, xpp, xppl, xppr;;) {
                if ((xp = x.parent) == null) {                         //如果插入的节点没有父节点, 就说明插入了根节点, 变为黑
                    x.red = false;
                    return x;
                }
                else if (!xp.red || (xpp = xp.parent) == null)        //插入的节点父节点是黑的, 或者x的爷爷是空, 直接插入即可
                    return root;                                      //
                if (xp == (xppl = xpp.left)) {                        //下面开始重点了, xp是红, 并且是左孩子, xppr存在并且是红
                    if ((xppr = xpp.right) != null && xppr.red) {     //            xpp(黑)                    xpp(红)            (这时候会问, 把xpp弄成红的了, 如果xppp是红的, 不就矛盾了吗)
                        xppr.red = false;                             //           /      \                  /      \             (事实上, 开头有个for循环, 就是不断处理这种问题的, x = xpp就是这个循环的更新)
                        xp.red = false;                               //         xp(红)   xppr(红)  -->  xp(黑)   xppr(黑)
                        xpp.red = true;                               //        /                         /
                        x = xpp;                                      //       x(红)                    x(红)
                    }
                    else {
                        if (x == xp.right) {                                  //否则就要旋转了
                            root = rotateLeft(root, x = xp);                       //如果x是一个右孩子, 先把他弄成左孩子的父亲然后x指向旋转后的xp, 然后交换xp和xpp的颜色, 再旋转调成平衡. 如果x是左孩子, 直接旋转平衡即可   
                            xpp = (xp = x.parent) == null ? null : xp.parent;         //  xpp(黑)                 xpp(黑)              xpp(黑)                  xpp(红)                    xp(黑)
                        }                                                            //  /     \                  /   \                /    \                 /      \                    /     \
                        if (xp != null) {                                           //  xp(红)  xppl(?)  ->     x(红) xppl(?)  ->   xp(红)   xppl(?)   -->   xp(黑)   xppl(?)     -->    x(红)  xpp(红)           (在每个for循环中, x的深度就减少1, 直到根节点, 或者中途return)
                            xp.red = false;                                        //  /   \                    /                   /                        /                           /         \
                            if (xpp != null) {                                    //xpl(?)  x(红)              xp(红)              x(红)                    x(红)                       xpl(?)      xppl(?)
                                // xpp.red = true;                                  //                        /                   /                         /
                                root = rotateRight(root, xpp);                  //                          xpl(?)              xpl(?)                    xpl(?)                 
                            } 
                        }
                    }
                }
                else {                                                // xp是右孩子的情况, 和上面的讨论类似, 就不写了, 画个图思路就清晰很多了. (手画二叉树实在是太麻烦了)
                    if (xppl != null && xppl.red) {
                        xppl.red = false;
                        xp.red = false;
                        xpp.red = true;
                        x = xpp;
                    }
                    else {
                        if (x == xp.left) {
                            root = rotateRight(root, x = xp);
                            xpp = (xp = x.parent) == null ? null : xp.parent;
                        }
                        if (xp != null) {
                            xp.red = false;
                            if (xpp != null) {
                                xpp.red = true;
                                root = rotateLeft(root, xpp);
                            }
                        }
                    }
                }
            }
        }

        static <K,V> TreeNode<K,V> balanceDeletion(TreeNode<K,V> root,          //这个函数是修复删节点之后的红黑树的, 而不是删除节点
                                                   TreeNode<K,V> x) {
                                                                               //整个函数是一个循环过程，可能会经过若干次循环。不管是刚调用此函数的第一次循环，或者是以后的循环，
            for (TreeNode<K,V> xp, xpl, xpr;;) {                              //每次循环体刚开始时，x节点子树的黑节点数，肯定是比x的兄弟节点子树的黑节点数少1，这是由removeTreeNode函数来做保证的（由于删掉了一个黑色节点，所以黑节点数少1）
                if (x == null || x == root)                                  //既然知道了x的黑节点数，比x的兄弟节点饿黑节点数少1，那么就需要通过调整来使得平衡。
                    return root;
                else if ((xp = x.parent) == null) {            // 边界情况, 可以直接返回. (x是root, x是red) x是red直接把x变黑就可以了
                    x.red = false;
                    return x;
                }
                else if (x.red) {
                    x.red = false;
                    return root;
                }
                else if ((xpl = xp.left) == x) {                   // x是左孩子
                    if ((xpr = xp.right) != null && xpr.red) {        //x的兄弟非空, 并且是红色的
                        xpr.red = false;                                          //      xp(黑)                    xp(红)                xpr(黑)
                        xp.red = true;                                           //      /    \         ->         /    \        ->      /     \
                        root = rotateLeft(root, xp);                            // x(黑,n-1)  xpr(红)        x(黑,n-1)  xpr(黑)         xp(红)  xprr(n)        至此只需要调节xp下面的两个节点不平衡的问题了, 进入下面的if
                        xpr = (xp = x.parent) == null ? null : xp.right;       //              / \                     /  \             /    \
                    }                                                         //         xprl(n)  xprr(n)         xprl(n)  xprr(n)  x(黑,n-1) xprl(黑,n) 
                    if (xpr == null)                                         //括号中的n代表从当前节点走到尽头需要经历n个黑色节点. 从一开始c的子树就比兄弟子树的黑节点少1.
                        x = xp;
                    else {
                        TreeNode<K,V> sl = xpr.left, sr = xpr.right;             // 剩下的不想画图了, 分类太多了, 在纸上画一画就出来了, 或者参考 https://blog.csdn.net/anlian523/article/details/103649200             
                        if ((sr == null || !sr.red) &&                          //即    xp(红)            这种情况的处理办法. 要对xpr的子节点进行分类讨论
                            (sl == null || !sl.red)) {                         //      /     \
                            xpr.red = true;                                   //  x(黑,n-1)  xpr(黑,n)
                            x = xp;
                        }
                        else {
                            if (sr == null || !sr.red) {
                                if (sl != null)
                                    sl.red = false;
                                xpr.red = true;
                                root = rotateRight(root, xpr);
                                xpr = (xp = x.parent) == null ?
                                    null : xp.right;
                            }
                            if (xpr != null) {
                                xpr.red = (xp == null) ? false : xp.red;
                                if ((sr = xpr.right) != null)
                                    sr.red = false;
                            }
                            if (xp != null) {
                                xp.red = false;
                                root = rotateLeft(root, xp);
                            }
                            x = root;
                        }
                    }
                }
                else { // symmetric                    //x是右孩子, 和上面的对称着来就可以了
                    if (xpl != null && xpl.red) {
                        xpl.red = false;
                        xp.red = true;
                        root = rotateRight(root, xp);
                        xpl = (xp = x.parent) == null ? null : xp.left;
                    }
                    if (xpl == null)
                        x = xp;
                    else {
                        TreeNode<K,V> sl = xpl.left, sr = xpl.right;
                        if ((sl == null || !sl.red) &&
                            (sr == null || !sr.red)) {
                            xpl.red = true;
                            x = xp;
                        }
                        else {
                            if (sl == null || !sl.red) {
                                if (sr != null)
                                    sr.red = false;
                                xpl.red = true;
                                root = rotateLeft(root, xpl);
                                xpl = (xp = x.parent) == null ?
                                    null : xp.left;
                            }
                            if (xpl != null) {
                                xpl.red = (xp == null) ? false : xp.red;
                                if ((sl = xpl.left) != null)
                                    sl.red = false;
                            }
                            if (xp != null) {
                                xp.red = false;
                                root = rotateRight(root, xp);
                            }
                            x = root;
                        }
                    }
                }
            }
        }

        /**
         * Recursive invariant check
         */
        static <K,V> boolean checkInvariants(TreeNode<K,V> t) {           //查看红黑树是否还是合法的
            TreeNode<K,V> tp = t.parent, tl = t.left, tr = t.right,      
                tb = t.prev, tn = (TreeNode<K,V>)t.next;
            if (tb != null && tb.next != t)                             //查看双向链表是不是正确的
                return false;
            if (tn != null && tn.prev != t)
                return false;
            if (tp != null && t != tp.left && t != tp.right)            //查看二叉树是不是正确的
                return false;
            if (tl != null && (tl.parent != t || tl.hash > t.hash))     //查看是不是BST
                return false;
            if (tr != null && (tr.parent != t || tr.hash < t.hash))
                return false;
            if (t.red && tl != null && tl.red && tr != null && tr.red)  //查看红黑节点是不是正确
                return false;
            if (tl != null && !checkInvariants(tl))                    //递归地查询左右子树
                return false;
            if (tr != null && !checkInvariants(tr))
                return false;
            return true;
        }
    }

}