关于java:Java-HashMap-源码

30次阅读

共计 4628 个字符,预计需要花费 12 分钟才能阅读完成。

HashMap 源码

属性

默认长度

如不传入初始化长度,则默认长度为 16

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

阈值

所能包容的 key-value 对的极限,超过此值就须要进行扩容

/**
 * The next size value at which to resize (capacity * load factor).
*/
int threshold;

底层数组

HashMap 底层寄存 Node 节点的数组,在第一次应用的时候进行初始化,长度总为 2 的 N 次幂

HashMap 保障扩容后长度 n 总为 2 次方是因为计算 Node 所在索引时采纳了 (n - 1) & hash 运算进行优化(& 比 % 效率更高),等价于对 n 取模,也就是 h % n

/**
 * 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;

负载因子

如不传入负载因子,默认为 0.75

负载因子过大会导致空间利用率较低,过小会导致碰撞概率变大,查问效率变低

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

扩容为树的阈值

当链表长度大于 8 时,会转变为红黑树

/**
 * 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;

回退为链表的阈值

当树中节点少于 6 个时,会转变为链表

/**
 * 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;

能够树化的最小底层数组长度

如果底层数组长度小于 64 时,阐明底层元素并不多,只是调配到某个地位的元素较多,先不转变为红黑树,先扩容底层数组扩散开

/**
 * 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;

treeifyBin(Node<K,V>[] tab, int hash) 办法中,如果当先底层数组长度小于 MIN_TREEIFY_CAPACITY 会进行扩容

if (tab == null || (n = tab.length) < MIN_TREEIFY_CAPACITY)
    resize();

重要办法

hash 算法

采纳 hashCode() 的高 16 位异或低 16 位实现,能够保障高下 16 位都会参加到运算中

static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}

put 流程

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
               boolean evict) {Node<K,V>[] tab; Node<K,V> p; int n, i;
    // 第一次应用,底层数组还没有初始化,进行初始化
    if ((tab = table) == null || (n = tab.length) == 0)
        n = (tab = resize()).length;
    // n 为底层数组长度,(n - 1) & hash 后果为 [0,n-1]
    // 如果为 null 则此处不存在任何 Node,间接新建 Node 放到此处即可
    // HashMap 保障扩容后长度 n 总为 2 次方,因为 & 比 % 具备更高的效率
    // 所以采纳了 (n - 1) & hash 运算进行优化,等价于对 n 取模,也就是 h % n
    if ((p = tab[i = (n - 1) & hash]) == null)
        tab[i] = newNode(hash, key, value, null);
    else {
        Node<K,V> e; K k;
        // 判断 Key 是否曾经存在,如果 key 存在,间接解决抵触
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            e = p;
        // 曾经转化为红黑树了
        else if (p instanceof TreeNode)
            // 把节点插入到红黑树中
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        else { // 仍是链表,采纳拉链法解决抵触
            for (int binCount = 0; ; ++binCount) {
                // 遍历到了链表尾部,阐明链表中 key 不存在抵触
                if ((e = p.next) == null) {
                    // 把新的 Node 插入到链表尾部
                    p.next = newNode(hash, key, value, null);
                    // 在链表尾部新插入了一个 Node,此时链表长度为 binCount + 1
                    // 相当于:链表长度 (binCount + 1) >= TREEIFY_THRESHOLD
                    if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                        // 转变为红黑树
                        treeifyBin(tab, hash);
                    break;
                }
                // 链表内有存在 Key,跳出循环,解决抵触
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                p = e;
            }
        }
        // 解决 key 抵触的状况
        if (e != null) { // existing mapping for key
            V oldValue = e.value;
            if (!onlyIfAbsent || oldValue == null)
                // 把抵触的 Node 的 value 替换掉
                e.value = value;
            afterNodeAccess(e);
            // 返回被替换的 Node 的 value
            return oldValue;
        }
    }
    ++modCount;
    // 超过能包容 KV 对的阈值,进行扩容
    if (++size > threshold)
        resize();
    afterNodeInsertion(evict);
    return null;
}

扩容办法

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) {
        // 超过最大值,不再进行扩容,此时会产生大量碰撞
        if (oldCap >= MAXIMUM_CAPACITY) {
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }
        // 扩容为原来的两倍
        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);
    }
    // 计算新的 KV 对阈值
    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) { // 从新计算索引,放入到新表中
        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;
                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;
}

正文完
 0