HashMap源码笔记

1.构造函数
//默认size=16,默认负载因子0.75

1.1

public HashMap() {
    this.loadFactor = DEFAULT_LOAD_FACTOR;// all other fields defaulted
}

1.2

//map容量,负载因子

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);

}
/**

* Returns a power of two size for the given target capacity.

* 为了控制hashmap的容量一定是2的n次方

* 通过位移运算,找到大于或等于 cap 的 最小2的n次幂。位运算的速度要优于乘除加减

*/

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.put数据
2.1

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

2.2

static final int hash(Object key) {
        int h;
        // h >>> 16 为了让高位参与运算,使得更均匀
        // (h = key.hashCode()) ^ (h >>> 16)
        // 先计算key的hash值,然后再用哈希值与自己右移16位做异或运算
        // 使用异或 ^运算符 使得运算结果更有随机性
        // PS:因为或运算,值会更偏向于1,与运算,值会更偏向于0

        return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
    }

2.3

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
        Node[] tab; Node p; int n, i;
        //第一次put数据的时候,hashmap才会初始化,并不是在构造函数中
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        //由于容量是2的n次方,所以 (n - 1) & hash 其实等同于 n % hash
        //而且这样运算的效率会更高
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
            Node 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)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;
    }

2.4

final Node[] resize() {
        Node[] oldTab = table;
        int oldCap = (oldTab == null) ? 0 : oldTab.length;
        //旧的扩容阈值
        int oldThr = threshold;
        //新的容量和扩容阈值
        int newCap, newThr = 0;
        //说明已经是旧的hashmap可进行resize
        if (oldCap > 0) {
            //hashmap >=最大容量
            if (oldCap >= MAXIMUM_CAPACITY) {
                //扩容阈值=Integer.max
                threshold = Integer.MAX_VALUE;
                return oldTab;
            }
            //容量扩容1倍 后小于最大容量 且  老容量 >= 初始容量
            //为什么扩容一倍大小呢?
            //扩容一倍大小的原因是
            // (1)为了保证hash 到数组位置的效率
            // (2)关系到扩容后元素在newCap中的放置问题:
            else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                     oldCap >= DEFAULT_INITIAL_CAPACITY)
                //扩容阈值也扩大1倍
                newThr = oldThr << 1; // double threshold
        }
        else if (oldThr > 0) // initial capacity was placed in threshold
            newCap = oldThr;
        else {
            // firstTime: 第一次初始化的时候,并且是没有入参的构造函数,会走进这里
            // zero initial threshold signifies using defaults
            // 如果是new 了一个新的hashmap的时候,会走这一行,初始化容量和扩容阈值
            newCap = DEFAULT_INITIAL_CAPACITY;
            newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
        }
        //firstTime第一次初始化的时候不是0了已经
        if (newThr == 0) {
            float ft = (float)newCap * loadFactor;
            newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
                      (int)ft : Integer.MAX_VALUE);
        }
        //设置hashmap长度阈值
        threshold = newThr;
        @SuppressWarnings({"rawtypes","unchecked"})
            Node[] newTab = (Node[])new Node[newCap];
        table = newTab;


        //如果不是初始化进来的
        if (oldTab != null) {
            //要遍历所有的节点
            for (int j = 0; j < oldCap; ++j) {
                Node e;
                //节点为null就continue 不为空的时候,把节点给临时节点e
                if ((e = oldTab[j]) != null) {
                    //把老结点 置为null
                    oldTab[j] = null;
                    //如果该结点没有next,也就是没有发生过冲突,桶的位置只有一个元素,把节点放入新的table中
                    if (e.next == null)
                        //就把值直接扔到对应的位置。这里有个问题,没有发生过冲突的,就直接放进去了,扩容后也不会发生冲突吗
                        // ,感觉这个很nb啊
                        newTab[e.hash & (newCap - 1)] = e;
                    //如果节点发生过冲突,并且是红黑树的数据结构
                    else if (e instanceof TreeNode)
                        //说明冲突的个数大于8个,那么就把树切割开
                        ((TreeNode)e).split(this, newTab, j, oldCap);
                    else { // preserve order
                        Node loHead = null, loTail = null;
                        Node hiHead = null, hiTail = null;
                        Node 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;
    }

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