HashMap源码解析——基于jdk11

HashMap源码解析——基于jdk11

插入流程

首先我们从put方法说起:

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

继续往下看putVal方法:

final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
                   boolean evict) {
                   
        //如果table为空就初始化一个table n为table的长度 
        Node<K,V>[] tab; Node<K,V> p; int n, i;
        if ((tab = table) == null || (n = tab.length) == 0)
            n = (tab = resize()).length;
        //如果数据将要存放在数组的位置为空 那么就把该值的放在数组上 
        if ((p = tab[i = (n - 1) & hash]) == null)
            tab[i] = newNode(hash, key, value, null);
        else {
        	//如果key已经存在, e = 存在的值
            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) {
                	//正常情况下,把值插入链表的最后面
                    if ((e = p.next) == null) {
                        p.next = newNode(hash, key, value, null);
                       //如果一个节点的元素太多 这个节点使用红黑树
                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st
                            treeifyBin(tab, hash);
                        break;
                    }
                    //如果key已经存在, e = 存在的值
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        break;
                    p = e;
                }
            }
            //e != null,说明存在相同的key
            if (e != null) { // existing mapping for key
                V oldValue = e.value;
                //在一定条件下新的值可以覆盖旧的值
                if (!onlyIfAbsent || oldValue == null)
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;
            }
        }
        ++modCount;
        //数据超过一定的数量 扩容
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
        return null;
    }

获取流程

首先看看

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

然后我们再看

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 &&
        	//找到链表的头
            (first = tab[(n - 1) & hash]) != null) {
            //如果链表头就是要找的key 返回结果
            if (first.hash == hash && // always check first node
                ((k = first.key) == key || (key != null && key.equals(k))))
                return first;
            //如果不是 就遍历查找链表
            if ((e = first.next) != null) {
                if (first instanceof TreeNode)
                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);
                do {
                    if (e.hash == hash &&
                        ((k = e.key) == key || (key != null && key.equals(k))))
                        return e;
                } while ((e = e.next) != null);
            }
        }
        return null;
    }
    原文作者:meak_962
    原文地址: https://blog.csdn.net/meak_962/article/details/83020863
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
点赞