关于这篇文章的 Volatile 问题

Question

我看到了这个：

Pattern #2: one-time safe publication
The visibility failures that
are possible in the absence of synchronization can get even trickier
to reason about when writing to object references instead of primitive
values. In the absence of synchronization, it is possible to see an
up-to-date value for an object reference that was written by another
thread and still see stale values for that object's state. (This
hazard is the root of the problem with the infamous
double-checked-locking idiom, where an object reference is read
without synchronization, and the risk is that you could see an
up-to-date reference but still observe a partially constructed object
through that reference.)*

来自 IBMdeveloperworks_volatile

我很困惑：

1. 什么是“原始值”
2. 原因：'看到另一个线程写入的对象引用的最新值，但仍然看到该对象状态的陈旧值'

Answer 1

1) 原始值不是像 int、boolean、float 等对象。请记住，存在与这些对象等价的对象，例如 Integer、Boolean 等。

2) 此线程可能会看到更新的对象引用，但该对象可能未完全初始化。例如，如果当此线程获取对象的新更新引用时构造函数尚未完成，则可能会发生这种情况......这可能很复杂，但可能会发生在某些 JVM 实现中。