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不知道为什么有这么多人在测试 volatile 的作用的时候喜欢用 println 这个方法，然后因为 println 方法的特殊性就得出一个结论，volatile 其实加和不加都一样。 首先，println 内部有 synchronized ，它的特殊性指的就是这一点。这会导致缓存失效，进而导致变量的重新读取，无论加没加 volatile 。
while(aBoolean){ System.out.println("running"); } 上面这种方式检验 volatile 是绝对错误的！ 正确的方式是
while(aBoolean){} System.out.println("done"); 这将导致线程永远空转，而加了 volatile 之后就可以正确的输出 done
 | 1 L4Linux Dec 4, 2022 via Android println 的 synchronized 只保证输出的原子性吧，管得着方法外的变量吗？你不妨举例说说其它 threads 有怎样的代码。 |
 | 2 wangyu17455 OP @L4Linux Concurrent Programming in Java 里说了啊 In essence, releasing a lock forces a flush of all writes from working memory employed by the thread, and acquiring a lock forces a (re)load of the values of accessible fields. While lock actions provide exclusion only for the operations performed within a synchronized method or block, these memory effects are defined to cover all fields used by the thread performing the action. 是当前作用域的所有变量都 reload from memory |
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