ArrayList不安全的实例及解决方案

public class ListTest {
    public static void main(String[] args) {
        List<String> list = Collections.synchronizedList(new ArrayList<>());
//        List<String> list = new Vector<>();
//        List<String> list = Collections.synchronizedList(new ArrayList<>());
        for (int i = 0; i < 20; i++) {
            new Thread(()->{
                list.add(UUID.randomUUID().toString().substring(0,8));
                list.forEach(System.out::println);
            }).start();
        }
    }
}

    public boolean add(E e) {
        // 加锁
        final ReentrantLock lock = this.lock;
        lock.lock();
        try {
            Object[] elements = getArray();
            int len = elements.length;
            // 复制一个新数组
            Object[] newElements = Arrays.copyOf(elements, len + 1);
            newElements[len] = e;
            // 新数组写完成后将旧的引用指向新数组的引用
            setArray(newElements);
            return true;
        } finally {
            lock.unlock();
        }
    }

    final void setArray(Object[] a) {
        array = a;
    }

public class ListTest {
    public static void main(String[] args) {
        Set<String> set = new CopyOnWriteArraySet<>();
        for (int i = 0; i < 20; i++) {
            new Thread(()->{
                set.add(UUID.randomUUID().toString().substring(0,8));
                set.forEach(System.out::println);
            }).start();
        }
    }
}