java——详述HashSet类add方法(二)

主程序:

import java.util.HashSet;

public class Test2 {

	public static void main(String[] args) {
		HashSet<String> names = new HashSet<String>();
		names.add("Jim");
		names.add("Jim");//这篇讲解这行如何执行
	}
}

HashSet中的add方法:
在add方法的底层代码中,又调用了put方法

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

HashMap中的put方法:
在put中又调用了hash方法和putVal方法

public V put(K key, V value) {
	return putVal(hash(key), key, value, false, true);//因为存的第一个对象和第二个要存的对象的哈希值相同,所以第二次存的对象与第一次存的对象hash(key)结果是一样的
}

hash方法:

static final int hash(Object key) {
     int h;
     return (key == null) ? 0 : (h = key.hashCode()) ^ (h >>> 16);//key不为零时,返回key的哈希值
}

HashMap中的putVal方法:
又调用了resize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)//由于存第一个对象时table全局变量已经有值了,所以table,也就是tab不为空,而且tab.length不等于0,数组的长度为16,所以该if为false,所以这个if语句块不执行
            n = (tab = resize()).length;
        if ((p = tab[i = (n - 1) & hash]) == null)//因为第一个和第二个值的hashCode相同,所以存第一个值和存第二个值的时候下标的值是一样的,因为第一次已经在此(n - 1) & hash 位置存了数据,所以tab[i = (n - 1) & hash]不为null,也就是p不为null,所以该if为false,所以这个if语句块不执行,将执行else语句块  结论:向HaspMap存储数据时,如果当前数据的hashcode和集合中已存在的元素的hashcode相同,则该if不成立
            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))))//p.hash为第一个Jim的hash值,所以p.hash == hash 成立,由于两次的对象时一个地址,所以(k = p.key) == key也成立,【当是names.add(new String("Jim"));时 则(p.hash == hash &&((k = p.key) == key)为false;而(key != null && key.equals(k)) 为true】
                e = p;//将第一个Jim的地址赋给e
            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;// e.value  常量Object
                if (!onlyIfAbsent || oldValue == null)// 在put方法中传给putVal方法的onlyIfAbsent本来就为false,所以!onlyIfAbsent true
                    e.value = value;
                afterNodeAccess(e);
                return oldValue;//返回第一个Jim,存储失败
            }
        }
        ++modCount;
        if (++size > threshold)
            resize();
        afterNodeInsertion(evict);
}

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];
        table = newTab;//将newTab赋给table
        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;//resize方法返回newTab
    }
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转载自blog.csdn.net/qq_44687512/article/details/105666082