HashMap源码分析总结

2019-12-11  本文已影响0人  一个忙来无聊的人

以下内容为jdk1.8版本。
声明:部分内容摘抄至网上,部分内容为个人见解,如有不对地方欢迎指正,蟹蟹
部分地方没有深入研究。

一、开篇

HashMap实现了java.lang的cloneable方法和序列化

public class HashMap<K,V> extends AbstractMap<K,V> implements Map<K,V>, Cloneable, Serializable {...}
//public abstract class AbstractMap<K,V> implements Map<K,V> {...}

1.1 、 HashMap 的默认长度 1 << 4 也就是2的4次方 16

我们都知道桶数组如果扩容,会申请内存空间,然后把原桶中的元素复制进新的桶数组中,这是一个比较耗时的过程。既然这样,那为何不把这两个值都设置大一些呢,设置大一些就不那么容易会进行扩容了啊。如果桶初始化桶数组设置太大,就会浪费内存空间,16是一个折中的大小,既不会像1,2,3那样放几个元素就扩容,也不会像几千几万那样可以只会利用一点点空间从而造成大量的浪费。

1.2、 HashMap 的最大长度 1 << 30 也就是2的30次方,同时也是 (int的最大值+1)/2 = 1073741824

···

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;

···
解释:由于HashMap规定了其容量是2的n次方,所以我们采用位运算<<来控制HashMap的大小。使用位运算同时还提高了Java的处理速度。
HashMap内部由Entry[]数组构成,Java的数组下标是由Int表示的。所以对于HashMap来说其最大的容量应该是不超过int最大值的一个2的指数幂,
而最接近int最大值的2个指数幂用位运算符表示就是 1 << 30

1.3、默认负载因子 0.75。达到容量时按照当前容量扩容0.75倍进行扩容,也就是当前容量2 ,如初始大小为16,当容量达到12(160.75)时就进行扩容,扩容到大小为32(16*2)。然后重新计算每个元素在数组中的位置,而这是一个非常消耗性能的操作,所以如果我们已经预知hashmap中元素的个数,那么预设元素的个数能够有效的提高hashmap的性能

static final float DEFAULT_LOAD_FACTOR = 0.75f;

解释:
默认负载因子(0.75)在时间和空间成本上提供了很好的折衷。较高的值会降低空间开销,但提高查找成本(体现在大多数的HashMap类的操作,包括get和put)。设置初始大小时,应该考虑预计的entry数在map及其负载系数,并且尽量减少rehash操作的次数。如果初始容量大于最大条目数除以负载因子,rehash操作将不会发生。

1.4、树形阈值与树形最小容量

static final int TREEIFY_THRESHOLD = 8;  // 树形阈值
static final int UNTREEIFY_THRESHOLD = 6; // 树转换最小值
static final int MIN_TREEIFY_CAPACITY = 64;//树形最小容量

解释:
结点冲突数达到8时,就会对hash表进行调整,如果table容量小于64,那么会进行table扩容,如果不小于64,那么会将冲突数达8那个单链表调整为红黑树
如果原先是红黑树的,resize后冲突结点数少于6了,就把红黑树恢复成单链表
如果table的容量少于64,那么即使冲突结点数达到TREEIFY_THRESHOLD后不会把该单链表调整成红黑树,而是将table扩容

1.5、一些局部变量信息

全局都会用到

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

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

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

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

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

二、方法探讨

2.1、初始容量大小计算

规则:初始容量的大小必须为2的n次方,且扩容后的size大于等于指定的size.
eg: 指定初始容量10,构造时实际容量为 16;指定初始容量16,构造初始容量16;指定初始容量17,构造初始容量为32;

如果扩容后大小小于0,则设置容量为1,如果扩容后容量大于设置的最大值MAXIMUM_CAPACITY 也就是1073741824,则返回HashMap最大容量
计算初始容量大小源码

   /**
     * Returns a power of two size for the given target capacity.
     */
    static final int tableSizeFor(int cap) {
        int n = cap - 1;
        n |= n >>> 1;
        n |= n >>> 2;
        n |= n >>> 4;
        n |= n >>> 8;
        n |= n >>> 16;
        return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
    }

2.2、构造函数

指定长度和负载系数

/**
     * Constructs an empty <tt>HashMap</tt> with the specified initial
     * capacity and load factor.
     *
     * @param  initialCapacity the initial capacity
     * @param  loadFactor      the load factor
     * @throws IllegalArgumentException if the initial capacity is negative
     *         or the load factor is nonpositive
     */
    public HashMap(int initialCapacity, float loadFactor) {
        if (initialCapacity < 0)
            throw new IllegalArgumentException("Illegal initial capacity: " +
                                               initialCapacity);
        if (initialCapacity > MAXIMUM_CAPACITY)
            initialCapacity = MAXIMUM_CAPACITY;
        if (loadFactor <= 0 || Float.isNaN(loadFactor))
            throw new IllegalArgumentException("Illegal load factor: " +
                                               loadFactor);
        this.loadFactor = loadFactor;
        this.threshold = tableSizeFor(initialCapacity);
    }

// 仅指定长度,使用默认的负载系数0.75

 /**
     * 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);
    }

// 无参构造函数

 /**
     * 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
    }

分析:
无参构造函数初始容量默认16,负载系数0.75
仅指定容量构造函数也使用默认负载系数0.75,同时调用两个参数构造函数。最终容量大小为 大于指定的容量最近的2的n次方。
指定容量大小和负载系数构造函数调用2.1初始容量大小计算规则计算,同时设置负载系数为入参负载系数。容量大小同上。

// 通过map构造函数

public HashMap(Map<? extends K, ? extends V> m) {
        this.loadFactor = DEFAULT_LOAD_FACTOR;
        putMapEntries(m, false);
    }

2.3、函数(基础)

// 通过指定map m创建HashMap hashMap
解释:若hashMap本身为空,则根据指定的map m的大小来初始化 hashMap大小
若hashMap为非空,则需要resize(),
最后再把key值和value值放进新map。
该方法在putAll方法中有使用到,

    /**
     * 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);
            }
        }
    }

//实现map的相关put方法
putVal在put元素时会先判断这个节点数组是否为空,如果为空的话初始化一个长度为16的数组。不为空的话继续判断经过&运算的索引的对象赋值给p,判断是否等于null。如果P为空直接新增。否则在后续代码块中会首先判断经过&运算的索引的对象p的key和要存入的对象的key是否相同,相同的话,把经过&运算的索引的对象p赋值到节点e中。不同的话判断p是否是树的实例,是的话就往树节点中put元素。通过for循环来put元素,这里也是解决hash冲突的地方。有的读者可能会问binCount是用来做什么的。binCount是一个记录这个for循环的次数,当已经获取完所有的元素时,会判断binCount是否>=7,是的话就转换成红黑树(转换规则:通过TREEIFY_THRESHOLD,和 UNTREEIFY_THRESHOLD 参数判断,也就是7及以上转换为红黑树,否则转换为链表)

 /**
     * Implements Map.put and related methods.
     *
     * @param hash hash for key
     * @param key the key
     * @param value the value to put
     * @param onlyIfAbsent if true, don't change existing value
     * @param evict if false, the table is in creation mode.
     * @return previous value, or null if none
     */
    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node<K,V> e; K k;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                e = p;
            else if (p instanceof TreeNode)
                e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
            else {
                for (int binCount = 0; ; ++binCount) {
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

put 的解释是:将指定的值和键相关联,若指定的键已经存在,则替换为新的值

    /**
     * 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);
    }

getNode方法是由两个值构成的,要存入的key的hash值和key,会先判断这个数组是不是为空,不为空的话,继续下面的判断,由于key的hash是可能大于数组的长度的,所以通过&运算保证的得出的索引值是小于数组的长度,并且将索引值的Node节点赋值给first,做key的判断相同则返回first,不相同的话则获取first的下一个节点判断是否为空,不为空的继续判断是否是树的实例,是的话去树节点中获取,不是的话就一直根据hash和key的判断,循环获取节点,直到获取到相同key的节点。
getNode方法在使用
判断是否包含key值
public boolean containsKey(Object key) {
return getNode(hash(key), key) != null;
}
获取指定的key值
public V get(Object key) {
Node<K,V> e;
return (e = getNode(hash(key), key)) == null ? null : e.value;
}

 /**
     * Implements Map.get and related methods.
     *
     * @param hash hash for key
     * @param key the key
     * @return the node, or null if none
     */
    final Node<K,V> getNode(int hash, Object key) {
        Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (first = tab[(n - 1) & hash]) != null) {
            if (first.hash == hash && // always check first node
                ((k = first.key) == key || (key != null && key.equals(k))))
                return first;
            if ((e = first.next) != null) {
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }

// 将链表里面的元素变成树形结构(除非表长度小于MIN_TREEIFY_CAPACITY也就是64)

/**
     * Replaces all linked nodes in bin at index for given hash unless
     * table is too small, in which case resizes instead.
     */
    final void treeifyBin(Node<K,V>[] tab, int hash) {
        int n, index; Node<K,V> e;
        if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
            resize();
        else if ((e = tab[index = (n - 1) & hash]) != null) {
            TreeNode<K,V> hd = null, tl = null;
            do {
                TreeNode<K,V> p = replacementTreeNode(e, null);
                if (tl == null)
                    hd = p;
                else {
                    p.prev = tl;
                    tl.next = p;
                }
                tl = p;
            } while ((e = e.next) != null);
            if ((tab[index] = hd) != null)
                hd.treeify(tab);
        }
    }

将链表结构转换为红黑树型结构

  /**
         * Forms tree of the nodes linked from this node.
         */
        final void treeify(Node<K,V>[] tab) {
            TreeNode<K,V> root = null;
            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) {
                    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;
                        else if (ph < h)
                            dir = 1;
                        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) {
                            x.parent = xp;
                            if (dir <= 0)
                                xp.left = x;
                            else
                                xp.right = x;
                            root = balanceInsertion(root, x);
                            break;
                        }
                    }
                }
            }
            moveRootToFront(tab, root);
        }

取消树化

 /**
         * Returns a list of non-TreeNodes replacing those linked from
         * this node.
         */
        final Node<K,V> untreeify(HashMap<K,V> map) {
            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;
        }

新增树节点和删除树节点

   /**
         * 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;;) {
                int dir, ph; K pk;
                if ((ph = p.hash) > h)
                    dir = -1;
                else if (ph < h)
                    dir = 1;
                else if ((pk = p.key) == k || (k != null && k.equals(pk)))
                    return p;
                else if ((kc == null &&
                          (kc = comparableClassFor(k)) == null) ||
                         (dir = compareComparables(kc, k, pk)) == 0) {
                    if (!searched) {
                        TreeNode<K,V> q, ch;
                        searched = true;
                        if (((ch = p.left) != null &&
                             (q = ch.find(h, k, kc)) != null) ||
                            ((ch = p.right) != null &&
                             (q = ch.find(h, k, kc)) != null))
                            return q;
                    }
                    dir = tieBreakOrder(k, pk);
                }

                TreeNode<K,V> xp = p;
                if ((p = (dir <= 0) ? p.left : p.right) == null) {
                    Node<K,V> xpn = xp.next;
                    TreeNode<K,V> x = map.newTreeNode(h, k, v, xpn);
                    if (dir <= 0)
                        xp.left = x;
                    else
                        xp.right = x;
                    xp.next = x;
                    x.parent = x.prev = xp;
                    if (xpn != null)
                        ((TreeNode<K,V>)xpn).prev = x;
                    moveRootToFront(tab, balanceInsertion(root, x));
                    return null;
                }
            }
        }

        /**
         * 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
         * 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,
                                  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;
            TreeNode<K,V> succ = (TreeNode<K,V>)next, pred = prev;
            if (pred == null)
                tab[index] = first = succ;
            else
                pred.next = succ;
            if (succ != null)
                succ.prev = pred;
            if (first == null)
                return;
            if (root.parent != null)
                root = root.root();
            if (root == null
                || (movable
                    && (root.right == null
                        || (rl = root.left) == null
                        || rl.left == null))) {
                tab[index] = first.untreeify(map);  // too small
                return;
            }
            TreeNode<K,V> p = this, pl = left, pr = right, replacement;
            if (pl != null && pr != null) {
                TreeNode<K,V> s = pr, sl;
                while ((sl = s.left) != null) // find successor
                    s = sl;
                boolean c = s.red; s.red = p.red; p.red = c; // swap colors
                TreeNode<K,V> sr = s.right;
                TreeNode<K,V> pp = p.parent;
                if (s == pr) { // p was s's direct parent
                    p.parent = s;
                    s.right = p;
                }
                else {
                    TreeNode<K,V> sp = s.parent;
                    if ((p.parent = sp) != null) {
                        if (s == sp.left)
                            sp.left = p;
                        else
                            sp.right = p;
                    }
                    if ((s.right = pr) != null)
                        pr.parent = s;
                }
                p.left = null;
                if ((p.right = sr) != null)
                    sr.parent = p;
                if ((s.left = pl) != null)
                    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)
                replacement = pl;
            else if (pr != null)
                replacement = pr;
            else
                replacement = p;
            if (replacement != p) {
                TreeNode<K,V> pp = replacement.parent = p.parent;
                if (pp == null)
                    root = replacement;
                else if (p == pp.left)
                    pp.left = replacement;
                else
                    pp.right = replacement;
                p.left = p.right = p.parent = null;
            }

            TreeNode<K,V> r = p.red ? root : balanceDeletion(root, replacement);

            if (replacement == p) {  // detach
                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);
        }

2.4、常用方法

remove和removeNode方法

    /**
     * Removes the mapping for the specified key from this map if present.
     *
     * @param  key key whose mapping is to be removed from the map
     * @return the previous value associated with <tt>key</tt>, or
     *         <tt>null</tt> if there was no mapping for <tt>key</tt>.
     *         (A <tt>null</tt> return can also indicate that the map
     *         previously associated <tt>null</tt> with <tt>key</tt>.)
     */
    public V remove(Object key) {
        Node<K,V> e;
        return (e = removeNode(hash(key), key, null, false, true)) == null ?
            null : e.value;
    }
 /**
     * Implements Map.remove and related methods.
     *
     * @param hash hash for key
     * @param key the key
     * @param value the value to match if matchValue, else ignored
     * @param matchValue if true only remove if value is equal
     * @param movable if false do not move other nodes while removing
     * @return the node, or null if none
     */
    final Node<K,V> removeNode(int hash, Object key, Object value,
                               boolean matchValue, boolean movable) {
        Node<K,V>[] tab; Node<K,V> p; int n, index;
        if ((tab = table) != null && (n = tab.length) > 0 &&
            (p = tab[index = (n - 1) & hash]) != null) {
            Node<K,V> node = null, e; K k; V v;
            if (p.hash == hash &&
                ((k = p.key) == key || (key != null && key.equals(k))))
                node = p;
            else if ((e = p.next) != null) {
                if (p instanceof TreeNode)
                    node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
                else {
                    do {
                        if (e.hash == hash &&
                            ((k = e.key) == key ||
                             (key != null && key.equals(k)))) {
                            node = e;
                            break;
                        }
                        p = e;
                    } while ((e = e.next) != null);
                }
            }
            if (node != null && (!matchValue || (v = node.value) == value ||
                                 (value != null && value.equals(v)))) {
                if (node instanceof TreeNode)
                    ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
                else if (node == p)
                    tab[index] = node.next;
                else
                    p.next = node.next;
                ++modCount;
                --size;
                afterNodeRemoval(node);
                return node;
            }
        }
        return null;
    }

请除方法
顾名思义,将指定的HashMap中的映射信息置空,并且将size设置为0.


    /**
     * 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;
        }
    }

验证是否包含指定key值

  /**
     * 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) {
                    if ((v = e.value) == value ||
                        (value != null && value.equals(v)))
                        return true;
                }
            }
        }
        return false;
    }
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