相关疑难解决方法(0)

鉴于以下multiton:

public class Multiton 
{
    private static final Multiton[] instances = new Multiton[...];

    private Multiton(...) 
    {
        //...
    }

    public static Multiton getInstance(int which) 
    {
        if(instances[which] == null) 
        {
            instances[which] = new Multiton(...);
        }

        return instances[which];
    }
}

如果没有getInstance()方法的昂贵同步和双重检查锁定的争议,我们如何保持线程安全和懒惰？这里提到了单身人士的有效方法,但似乎并没有扩展到多人.