详述HashSet类add方法（一）

import java.util.HashSet;
public class Test{
    public static void main(String[] args){
    HashSet<String> names = new HashSet<String>();//调用构造方法时，创建HashMap集合对象
    names.add("Jim");//向HashMap集合的key存值，HashMapvalue是一个常量Object
    }
}

在执行该代码时Jim为第一次存入

HashSet add方法代码：

public boolean add(E e){
    return map.put(e,PRESENT)==null;;
}

add方法中传入的e为Jim，PRESENT为HashMap的常量

HashMap put方法：

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

而put方法传入了Jim的putval方法返回值；在该方法中调用了hash方法

hash方法：

static final int hash(Object key){
    int h;
    return (key == null)?0:(h = key.hashCode())^(h>>>16);//当传入的key是null时将会返回0；当两个对象的hashCode相同时，他们的hash()返回值也是相同的
}

hashCode()方法：同一对象多次调用该方法结果都相同；当不同对象重写了该方法导致返回值相同，则多次调用该方法结果都相同

HashMap putVal方法：

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)//table为null，tab为null，则(tab = table) == null为true
            n = (tab = resize()).length;//resize()方法直接为table返回newTab，resize方法也返回newTab，所以，table与tab是一个对象，数组长度16
        if ((p = tab[i = (n - 1) & hash]) == null)// i = (n - 1) & hash 数组长度-1  & hash值，i是下表  tab[i = (n - 1) & hash]，因为目前只存储一个值，所以一定为null
            tab[i] = newNode(hash, key, value, null);// tab[i] I   (n - 1) & hash 为(n - 1) & hash添加一个元素，因为数组为地址传递，所以table全部变量i也有值
        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);

在调用putval方法时由于当前tab数组weinull所以通过调用resize方法创建了长度为16的tab[]数组；er由于wei第一次存值，所以tab[]数组中的值为null，此时将key值存入指定的位置中完成添加操作。

其中的resize()方法

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];//默认数组长度为16
        table = newTab;//将table赋值给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;//返回newTab
    }