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  • HashMap主要方法源码分析(JDK1.8)

    本篇从HashMap的put、get、remove方法入手,分析源码流程
    (不涉及红黑树的具体算法)
    jkd1.8中HashMap的结构为数组、链表、红黑树的形式
     
     
    (未转化红黑树时)
     

     (转化为红黑树时的情况)


    一、关于HashMap需要了解的静态常量

    DEFAULT_INITIAL_CAPACITY 数组默认初始容量 16
    DEFAULT_LOAD_FACTOR 默认负载因子 0.75
    MIN_TREEIFY_CAPACITY 最小树容量 64
    在下面的方法探究中将会提到这些静态常量的用处 
     

    二、方法探究

    1、put

    HashMap中的数组是第一次调用put方法时才创建对象的

    下面是从进入put到创建完数组的全过程

     

     将key、value作为参数传入后,计算key的hash,再传入putVal方法

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

    putVal()的整个流程如下        
    1、先判断table是否为null,如果是,调用resize()
     
    /**
         * Initializes or doubles table size.  If null, allocates in
         * accord with initial capacity target held in field threshold.
         * Otherwise, because we are using power-of-two expansion, the
         * elements from each bin must either stay at same index, or move
         * with a power of two offset in the new table.
         *
         * @return the table
         */
        final Node<K,V>[] resize() {
            Node<K,V>[] oldTab = table;
            int oldCap = (oldTab == null) ? 0 : oldTab.length;
            int oldThr = threshold;
            int newCap, newThr = 0;
            if (oldCap > 0) {
                if (oldCap >= MAXIMUM_CAPACITY) {
                    threshold = Integer.MAX_VALUE;
                    return oldTab;
                }
                else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                         oldCap >= DEFAULT_INITIAL_CAPACITY)
                    newThr = oldThr << 1; // double threshold
            }
            else if (oldThr > 0) // initial capacity was placed in threshold
                newCap = oldThr;
            else {               // zero initial threshold signifies using defaults
                newCap = DEFAULT_INITIAL_CAPACITY;
                newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
            }
            if (newThr == 0) {
                float ft = (float)newCap * loadFactor;
                newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
                          (int)ft : Integer.MAX_VALUE);
            }
            threshold = newThr;
            @SuppressWarnings({"rawtypes","unchecked"})
            Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
            table = newTab;
            if (oldTab != null) {
                for (int j = 0; j < oldCap; ++j) {
                    Node<K,V> e;
                    if ((e = oldTab[j]) != null) {
                        oldTab[j] = null;
                        if (e.next == null)
                            newTab[e.hash & (newCap - 1)] = e;
                        else if (e instanceof TreeNode)
                            ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                        else { // preserve order
                            Node<K,V> loHead = null, loTail = null;
                            Node<K,V> hiHead = null, hiTail = null;
                            Node<K,V> next;
                            do {
                                next = e.next;
                                if ((e.hash & oldCap) == 0) {
                                    if (loTail == null)
                                        loHead = e;
                                    else
                                        loTail.next = e;
                                    loTail = e;
                                }
                                else {
                                    if (hiTail == null)
                                        hiHead = e;
                                    else
                                        hiTail.next = e;
                                    hiTail = e;
                                }
                            } while ((e = next) != null);
                            if (loTail != null) {
                                loTail.next = null;
                                newTab[j] = loHead;
                            }
                            if (hiTail != null) {
                                hiTail.next = null;
                                newTab[j + oldCap] = hiHead;
                            }
                        }
                    }
                }
            }
            return newTab;
        }
    resize
    resize():
    判断当前table为null后,初始化负载因子DEFAULT_LOAD_FACTOR
    计算当前数组边界DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY
    (第一次数组边界为16*0.75=12,当数组超过数组边界时会扩大为两倍。
    也就是说数组中的元素达到或大于12时,将第一次扩大数组,大小变为16*2=32)
    创建并返回大小为DEFAULT_INITIAL_CAPACITY的table对象。
    (总的来说resize负责扩大数组容量和初始化数组)
    以上就是调用put方法时HashMap对象内数组的创建过程

    2、如果进入putVal()判断table不为null
    利用hash值与(&)计算出数组下标,并判断是否为空
    如果是,创建node对象并存入数组
    如果不是,从当前下标的链表第一位开始一个个往下对比
    ①若hash和key值都相同,则break退出循环,之后进行value的替换,并返回oldValue
    ②若过程中遇到null,则创建node对象
      判断是否达到红黑树转换条件:如果当前链表长度达到8,进入treeifyBin方法
      判断表长度(如果小于64,则调用resize(),判断要不要增大数组
      反之replacementTreeNode,用红黑树代替当前链表

    (向下遍历数组当前下标链表的操作)

    (key相同时value的替换操作)


    3、增加size(map中元素的数量)和modCount(对map的操作次数)

    2、get

    * 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;
        }
    get
    将key和key的hash传入getNode方法,并获取返回的Node对象的value
    先判断数组不为null,且长度大于0
    利用hash值先找到node可能在的列的第一个元素(当前传入key可能不存在)
    再竖向遍历链表对比hash和key值

    (上面的first即为node可能存在的链表的第一个元素) 

    查找时如果第一个即匹配则直接返回

     否则往下遍历直到匹配或node为null

    3、removeNode

    /**
         * 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;
        }
    removeNode

    与getNode的逻辑类似,利用hash值查找可能存在的位置

    如果链表第一位就匹配:

    对链表的遍历:

    判断node的类型,如果是链表则用node.nest来代替当前位置

    最后操作数+1,size-1,返回被remove的node对象

     
     
     
     
     
     
     
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  • 原文地址:https://www.cnblogs.com/Unicron/p/12721931.html
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