Hashmap 源码分析2

作者: 遗忘的流逝 | 来源:发表于2017-04-09 20:45 被阅读63次

    Hashmap 源码分析之基本操作

    我们从流程中看一下,这个hashmap究竟是这么实现的。
    我们先了解一下put方法实现

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

    接下来的才是真正实现:

    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
    boolean evict) {
    //数组tab table是内部数组
    Node<K,V>[] tab;
    Node<K,V> p;
    int n, i;
    if ((tab = table) == null || (n = tab.length) == 0)
    // resize()是扩容函数
    n = (tab = resize()).length;
    //取到hash地址,如果为null就可以填入
    if ((p = tab[i = (n - 1) & hash]) == null)
    tab[i] = newNode(hash, key, value, null);
    else {
    //不为null
    Node<K,V> e; K k;
    //如果map中已经存在,就将值覆盖
    if (p.hash == hash &&
    ((k = p.key) == key || (key != null && key.equals(k))))
    e = p;
    //如果这个结点是红黑树的结点,那么使用putTreeVal来实现。
    else if (p instanceof TreeNode)
    e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
    //如果是链表结构
    else {
    for (int binCount = 0; ; ++binCount) {
    if ((e = p.next) == null) {
    p.next = newNode(hash, key, value, null);
    //假如这个链表结点个数大于等于TREEIFY_THRESHOLD - 1的时候的时候,直接转换成树
    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
    treeifyBin(tab, hash);
    break;
    }
    // key已经存在直接覆盖value
    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;
    //onlyIfAbsent if true, don't change existing value
    if (!onlyIfAbsent || oldValue == null)
    e.value = value;

                afterNodeAccess(e);
                return oldValue;
            }
        }
    

    //这个是一个安全机制 modCount在LinkedList讲过
    ++modCount;
    if (++size > threshold)
    resize();
    afterNodeInsertion(evict);
    return null;
    }

    至于afterNodeInsertion这些,在hashmap中其实是没有实现的,

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

    正如上面写的,应该是和LinkedhashMap有关系。

    我们接下来看,get方法,注意hashmap是可以有null值的

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

    同样真正实现是在

    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) {
    //如果第一个存在,并可以key值相同返回first
    if (first.hash == hash && // always check first node
    ((k = first.key) == key || (key != null && key.equals(k))))
    return first;
    //没有找到的话
    if ((e = first.next) != null) {
    //先判断是否是红黑树
    if (first instanceof TreeNode)
    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
    do {
    //如果是链表开始判断
    if (e.hash == hash &&
    ((k = e.key) == key || (key != null && key.equals(k))))
    return e;
    } while ((e = e.next) != null);
    }
    }
    return null;
    }

    接下来就是hashmap扩容了

    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) {
    //如果超过了最大容量,不会扩容的
    if (oldCap >= MAXIMUM_CAPACITY) {
    threshold = Integer.MAX_VALUE;
    return oldTab;
    }
    //没有的话讲容量变成两倍
    else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
    oldCap >= DEFAULT_INITIAL_CAPACITY)
    newThr = oldThr << 1; // double threshold
    }
    //初始化
    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);
    }
    //重新计算下次扩容的临界值
    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) {
    //为null利于gc回收处理
    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);
    

    //将旧的table对于到扩容的table上
    //扩了一倍,那么其实是就是高位新增一位判断是1 还是0
    if (loTail != null) {
    loTail.next = null;
    newTab[j] = loHead;
    }
    if (hiTail != null) {
    hiTail.next = null;
    newTab[j + oldCap] = hiHead;
    }
    }
    }
    }
    }
    return newTab;
    }

    那么其实我们可以看到hashmap扩容其实非常的耗时间,如果可以预测这个map可以存的值数量,其实在初始化的时候定义是更加合算的。

    读源码,是我们了解大神领域的一大捷径
    生命不息,奋斗不止

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