HashMap的长度为什么要是2的N次方

旧城等待， 2023-07-07 12:37 7阅读 0赞

先贴一段源码压压惊，

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

见上面的源码中的注释，MUST be a power of two

我们知道hashmap是由数组和链表组成的，在Put进去一个值时，会先hashcode%tableLength,

而在计算机中位移运算是最快的，a%b=a-(a/b)\*b;而a%b=a&(b-1)的前提是b必须是2的N次幂。

见源码

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;
        }

其中i=(n-1)&hash就是上面介绍的原理.