HashMap的简单源码分析

关于HashMap

1.容量

2.加载因子

1和2的决定了方法resize()
并且加载因子会决定空间的利用率和hash冲突的几率，
默认的加载因子是0.75

    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) {            if (oldCap >= MAXIMUM_CAPACITY) {                threshold = Integer.MAX_VALUE;                return oldTab;            }            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);        }        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) {                    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;    }

3.Node

    static class Node<K,V> implements Map.Entry<K,V> {        final int hash;// 通过key对象产生的int值 具体参考hash方法        final K key;        V value;        // 链表        Node<K,V> next;        Node(int hash, K key, V value, Node<K,V> next) {            this.hash = hash;            this.key = key;            this.value = value;            this.next = next;        }        public final K getKey()        { return key; }        public final V getValue()      { return value; }        public final String toString() { return key + "=" + value; }        public final int hashCode() {            return Objects.hashCode(key) ^ Objects.hashCode(value);        }        public final V setValue(V newValue) {            V oldValue = value;            value = newValue;            return oldValue;        }        public final boolean equals(Object o) {            if (o == this)                return true;            if (o instanceof Map.Entry) {                Map.Entry<?,?> e = (Map.Entry<?,?>)o;                if (Objects.equals(key, e.getKey()) &&                    Objects.equals(value, e.getValue()))                    return true;            }            return false;        }    }

显而易见的链表

4.Node数组

put

    final V putVal(int hash, K key, V value, boolean onlyIfAbsent,                   boolean evict) {        Node<K,V>[] tab; Node<K,V> p; int n, i;        // 判断数组是否为空         if ((tab = table) == null || (n = tab.length) == 0)        // resize() 对数组进行初始化或者扩容 这个地方是初始化            n = (tab = resize()).length;            // 在数组中定位 hash 通过key对象产生的 具体参考-hash函数 如果该位置没有值        if ((p = tab[i = (n - 1) & hash]) == null)        // 为该位置赋值            tab[i] = newNode(hash, key, value, null);        else { // 当前位置有值            Node<K,V> e; K k;            // 判断当前位置上面的key和当前存放的key是否一致            if (p.hash == hash &&                ((k = p.key) == key || (key != null && key.equals(k))))                e = p; // key一致 进行值的覆盖            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) {                        // 这个地方体现出来了链表                        p.next = newNode(hash, key, value, null);                        if (binCount >= TREEIFY_THRESHOLD - 1) // -1 for 1st                        // 这个地方是一个优化的操作 如果连败哦的长度大于等于一个指定的值 则将链表改为树的结构                            treeifyBin(tab, hash);                        break;                    }                    // 判断是否key一致                    if (e.hash == hash &&                        ((k = e.key) == key || (key != null && key.equals(k))))                        break;                    p = e;                }            }            // 如果当前key已经存在进行值的覆盖            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) // 判断是否需要进行扩容            resize();        afterNodeInsertion(evict);        return null;    }

get

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

    final Node<K,V> getNode(int hash, Object key) {        Node<K,V>[] tab; Node<K,V> first, e; int n; K k;        // 关于first的赋值，与put相对应。找到每个数组位置对应的链表的第一个元素        if ((tab = table) != null && (n = tab.length) > 0 &&            (first = tab[(n - 1) & hash]) != null) {            if (first.hash == hash && // always check first node                ((k = first.key) == key || (key != null && key.equals(k))))                return first;            if ((e = first.next) != null) {                if (first instanceof TreeNode)                    return ((TreeNode<K,V>)first).getTreeNode(hash, key);                do { // 遍历链表                    if (e.hash == hash &&                        ((k = e.key) == key || (key != null && key.equals(k))))                        return e;                } while ((e = e.next) != null);            }        }        return null;    }

关于对链表的优化通过树实现请自行参考源码分析