CopyOnWriteArraySet源码分析

简介

内部基于CopyOnWriteArrayList实现的Set集合

类图

属性

	// 内部依赖CopyOnWriteArrayList实现
    private final CopyOnWriteArrayList<E> al;

构造方法

	// 内部使用CopyOnWriteArrayList存储数据
	public CopyOnWriteArraySet() {
        al = new CopyOnWriteArrayList<E>();
    }

    public CopyOnWriteArraySet(Collection<? extends E> c) {
        if (c.getClass() == CopyOnWriteArraySet.class) {
            @SuppressWarnings("unchecked") CopyOnWriteArraySet<E> cc =
                (CopyOnWriteArraySet<E>)c;
            al = new CopyOnWriteArrayList<E>(cc.al);
        }
        else {
            al = new CopyOnWriteArrayList<E>();
            al.addAllAbsent(c);
        }
    }

如何保证不重复

// 调用List的当不存在时添加方法
public boolean add(E e) {
    return al.addIfAbsent(e);
}
// 不存在才会添加
public boolean addIfAbsent(E e) {
    Object[] snapshot = getArray();
    return indexOf(e, snapshot, 0, snapshot.length) >= 0 ? false :
        addIfAbsent(e, snapshot);
}

private boolean addIfAbsent(E e, Object[] snapshot) {
    final ReentrantLock lock = this.lock;
    lock.lock();
    try {
        Object[] current = getArray();
        int len = current.length;
        if (snapshot != current) {
            // 说明在这之前已经被修改过了
            int common = Math.min(snapshot.length, len);
            // 判断元素是否存在于当前数组中，如果存在直接返回false;
            // 这里巧妙地思路：判断两个数组相等 1长度相等 2数组每一位相等 不需要进行双重循环
            for (int i = 0; i < common; i++)
                if (current[i] != snapshot[i] && eq(e, current[i]))
                    return false;
            if (indexOf(e, current, common, len) >= 0)
                    return false;
        }
        // 添加并修改
        Object[] newElements = Arrays.copyOf(current, len + 1);
        newElements[len] = e;
        setArray(newElements);
        return true;
    } finally {
        lock.unlock();
    }
}