HashMap(1.8)源码阅读

先了解一下用到的位运算符：https://www.cnblogs.com/gavinYang/p/11196492.html

一、初始化

1.无参构造函数：

//负载因子默认值
static final float DEFAULT_LOAD_FACTOR = 0.75f;

//指定loadFactor负载因子的值是0.75f
public HashMap() {
    this.loadFactor = DEFAULT_LOAD_FACTOR; // all other fields defaulted
}

2.指定初始化大小和负载因子：

//hashmap的最大容量
static final int MAXIMUM_CAPACITY = 1 << 30; //1左移30位等于1073741824

//initialCapacity:传入的hashmap初始化大小
//loadFactor:负载因子,此时是默认值0.75f
public HashMap(int initialCapacity, float loadFactor) {
    if (initialCapacity < 0)
        throw new IllegalArgumentException("Illegal initial capacity: " + initialCapacity);
    //如果大于最大容量，则初始化的值为最大容量的值
    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);
}

//返回一个大于等于initialCapacity的2的n次方最近的一个值
//假设传入的initialCapacity=5
static final int tableSizeFor(int cap) {
    int n = cap - 1;  //
    n |= n >>> 1; // 00000100 |= 00000010 得出 00000110=6
    n |= n >>> 2; // 00000110 |= 00000011 得出 00000111=7
    n |= n >>> 4; // 00000111 |= 00000000 得出 00000111=7
    n |= n >>> 8; // 00000111 |= 00000000 得出 00000111=7
    n |= n >>> 16;// 00000111 |= 00000000 得出 00000111=7
    //最终返回n+1=8
    return (n < 0) ? 1 : (n >= MAXIMUM_CAPACITY) ? MAXIMUM_CAPACITY : n + 1;
}

3.指定初始化大小（会调用2）

//initialCapacity：hashmap初始化大小
public HashMap(int initialCapacity) {
    this(initialCapacity, DEFAULT_LOAD_FACTOR);
}

二、put元素(转红黑树和put一个TreeNode时待补充)

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

//计算hash值
static final int hash(Object key) {
    int h;
    return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);
}
//hashmap的数组
transient Node<K,V>[] table;

//存放数据
final V putVal(int hash, K key, V value, boolean onlyIfAbsent,
               boolean evict) {
    Node<K,V>[] tab; Node<K,V> p; int n, i;
    //把table赋值给tab，判断tab是否为空(第一次put时执行)
    if ((tab = table) == null || (n = tab.length) == 0)
        //返回数组的长度
        n = (tab = resize()).length;
    //计算一个位置(& 运算后值肯定小于 n-1),如果数组的这个位置不为空则直接放入数据
    if ((p = tab[i = (n - 1) & hash]) == null)
        tab[i] = newNode(hash, key, value, null);
    //数组不为空，且计算的位置以及有数据
    else {
        Node<K,V> e; K k;
        //添加的数据与所在的数组的位置的hash值和key相同，则将当前数组位置的数据数据p赋值给e
        if (p.hash == hash &&
            ((k = p.key) == key || (key != null && key.equals(k))))
            e = p;
        //如果数组当前位置的值是TreeNode类型。。。。。。。
        else if (p instanceof TreeNode)
            e = ((TreeNode<K,V>)p).putTreeVal(this, tab, hash, key, value);
        else {
            //死循环，binCount表示链表长度
            for (int binCount = 0; ; ++binCount) {
                //判断当前数组位置是否有下一个值
                if ((e = p.next) == null) {
                    //数组位置的下一个值是当前新插入的数据(插入尾部)
                    p.next = newNode(hash, key, value, null);
                    //链表长度大于TREEIFY_THRESHOLD=8时转化为红黑树
                    if (binCount >= TREEIFY_THRESHOLD - 1)
                        treeifyBin(tab, hash);
                    break;
                }
                //添加的数据与如果下一个值的hash值和key相同，直接跳出
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    break;
                //下一个位置的数据赋值给p，循环继续
                p = e;
            }
        }
        //key存在，更新新值，返回旧值
        if (e != null) { // existing mapping for key
            V oldValue = e.value;
            if (!onlyIfAbsent || oldValue == null)
                e.value = value;
            afterNodeAccess(e);
            return oldValue;
        }
    }
    ++modCount;
    //判断是否超过需要的大小
    if (++size > threshold)
        resize();
    afterNodeInsertion(evict);
    return null;
}

//初始化或扩容时调用
final Node<K,V>[] resize() {
    //将原有的数据存放到oldTab
    Node<K,V>[] oldTab = table;
    //旧数组的容量
    int oldCap = (oldTab == null) ? 0 : oldTab.length;
    //需要扩容的大小
    int oldThr = threshold;
    //新的容量，新的需要扩容的大小
    int newCap, newThr = 0;
    //旧数组有数据
    if (oldCap > 0) {
        //旧数组容量大于等于最大容量1 << 30 = 1073741824
        if (oldCap >= MAXIMUM_CAPACITY) {
            //不扩容返回旧数组
            threshold = Integer.MAX_VALUE;
            return oldTab;
        }
        //新的容量是旧数组的容量左移以为即：oldCap*2。如果数组容量大于DEFAULT_INITIAL_CAPACITY=16时，扩容的容量也要×2
        else if ((newCap = oldCap << 1) < MAXIMUM_CAPACITY &&
                 oldCap >= DEFAULT_INITIAL_CAPACITY)
            newThr = oldThr << 1; // double threshold
    }
    //旧数组为空，进行初始化
    else if (oldThr > 0) // 初始化容量设置为threshold(需要扩容的值)
        newCap = oldThr;
    else {
        //无参初始化hashmap时，容量和扩容的大小
        newCap = DEFAULT_INITIAL_CAPACITY;
        newThr = (int)(DEFAULT_LOAD_FACTOR * DEFAULT_INITIAL_CAPACITY);
    }
    //新的threshold(需要扩容的大小)为零时,newThr=newCap * loadFactor
    if (newThr == 0) {
        float ft = (float)newCap * loadFactor;
        newThr = (newCap < MAXIMUM_CAPACITY && ft < (float)MAXIMUM_CAPACITY ?
                  (int)ft : Integer.MAX_VALUE);
    }
    //赋值回threshold
    threshold = newThr;
    @SuppressWarnings({"rawtypes","unchecked"})
    //定义一个新hashmap的数组，值是新的容量
    Node<K,V>[] newTab = (Node<K,V>[])new Node[newCap];
    table = newTab;
    //旧数组不为空，扩容
    if (oldTab != null) {
        //遍历旧的数组
        for (int j = 0; j < oldCap; ++j) {
            Node<K,V> e;
            if ((e = oldTab[j]) != null) {
                oldTab[j] = null;
                //数组当前位置下没有链表
                if (e.next == null)
                    //重新计算hash然后放入新数组的新位置
                    newTab[e.hash & (newCap - 1)] = e;
                else if (e instanceof TreeNode)
                    ((TreeNode<K,V>)e).split(this, newTab, j, oldCap);
                else {
                    //数组位置下有链表
                    Node<K,V> loHead = null, loTail = null;
                    Node<K,V> hiHead = null, hiTail = null;
                    Node<K,V> next;
                    do {
                        next = e.next;
                        //e.hash & oldCap 返回0或者oldCap
                        //如果链表中各节点(e.hash & oldCap)计算值不一样时数据会分布在数组的旧位置和(旧位置+oldCap)
                        if ((e.hash & oldCap) == 0) {
                            //第一次遍历时，loHead等于当前链表
                            //loTail每循环一次去掉链表头部
                            if (loTail == null)
                                loHead = e;
                            else
                                loTail.next = e;
                            loTail = e;
                        }
                        else {
                            //第一次遍历时，hiHead等于当前链表
                            //hiTail每循环一次去掉链表头部
                            if (hiTail == null)
                                hiHead = e;
                            else
                                hiTail.next = e;
                            hiTail = e;
                        }
                    } while ((e = next) != null);
                    //将当前数组位置下的链表(hiHead/loHead)放到新数组的(当前数组位置+oldCap)位置
                    //loTail/hiTail.next设置为空为了防止loTail和loHead(或hiHead和hiTail)指向同一个地址时数据重复，
                    //重复时则清除loHead和hiHead的next值
                    if (loTail != null) {
                        loTail.next = null;
                        newTab[j] = loHead;
                    }
                    if (hiTail != null) {
                        hiTail.next = null;
                        newTab[j + oldCap] = hiHead;
                    }
                }
            }
        }
    }
    return newTab;
}

三、get元素

//返回键对应的值
public V get(Object key) {
    Node<K,V> e;
    return (e = getNode(hash(key), key)) == null ? null : e.value;
}

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

//根据key获得对应的node
final Node<K,V> getNode(int hash, Object key) {
    Node<K,V>[] tab; Node<K,V> first, e; int n; K k;
    //数组不等于空，要获取的元素的位置不等于空
    if ((tab = table) != null && (n = tab.length) > 0 && (first = tab[(n - 1) & hash]) != null) {
        //第一个元素即是要查找的元素
        if (first.hash == hash && ((k = first.key) == key || (key != null && key.equals(k))))
            return first;
        //获取链表的下一个，准备开始遍历链表
        if ((e = first.next) != null) {
            if (first instanceof TreeNode)
                return ((TreeNode<K,V>)first).getTreeNode(hash, key);
            //遍历链表
            do {
                if (e.hash == hash &&
                    ((k = e.key) == key || (key != null && key.equals(k))))
                    return e;
            } while ((e = e.next) != null);
        }
    }
    return null;
}

四、remove元素

//返回键对应的值
public V remove(Object key) {
    Node<K,V> e;
    return (e = removeNode(hash(key), key, null, false, true)) == null ?
        null : e.value;
}

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

//value：如果是matchValue则需要传入
//matchValue：如果等于true则还需要匹配值也相等
//movable：如果为false则在删除时不要移动其他节点
final Node<K,V> removeNode(int hash, Object key, Object value,boolean matchValue, boolean movable) {
    Node<K,V>[] tab; Node<K,V> p; int n, index;
    //数组不等于空，要获取的元素的数组位置不等于空
    if ((tab = table) != null && (n = tab.length) > 0 && (p = tab[index = (n - 1) & hash]) != null) {
        Node<K,V> node = null, e; K k; V v;
        //如果数组当前位置p即是对应的值
        if (p.hash == hash && ((k = p.key) == key || (key != null && key.equals(k))))
            node = p;
        //获取链表的下一个，准备开始遍历链表
        else if ((e = p.next) != null) {
            if (p instanceof TreeNode)
                node = ((TreeNode<K,V>)p).getTreeNode(hash, key);
            //遍历链表
            else {
                do {
                    if (e.hash == hash && ((k = e.key) == key || (key != null && key.equals(k)))) {
                        node = e;
                        break;
                    }
                    p = e;
                } while ((e = e.next) != null);
            }
        }
        //找到元素，并且matchValue=false或者找到的node的value与指定value相等
        if (node != null && (!matchValue || (v = node.value) == value || (value != null && value.equals(v)))) {
            if (node instanceof TreeNode)
                ((TreeNode<K,V>)node).removeTreeNode(this, tab, movable);
            //如果节点在数组上，则直接将数组的位置指定为p的下一个位置
            else if (node == p)
                tab[index] = node.next;
            //如果节点在链表上，则将当前节点的下一个值指定为找到的节点的下一个值
            else
                p.next = node.next;
            ++modCount;
            --size;
            afterNodeRemoval(node);
            return node;
        }
    }
    return null;
}