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    <p>I have just found this question because I was asking myself the same thing. According to <a href="https://blogs.oracle.com/dave/entry/a_race_in_locksupport_park" rel="nofollow">this article</a> by Oracle researcher <a href="https://labs.oracle.com/pls/apex/f?p=labs%3abio:0:29" rel="nofollow">David Dice</a>, the answer seems to be <strong>no</strong>. Here's the relevant part of the article:</p> <blockquote> <p>If a thread is blocked in <code>park()</code> we're guaranteed that a subsequent <code>unpark()</code> will make it ready. A perfectly legal but low-quality implementation of <code>park()</code> and <code>unpark()</code> would be empty methods, in which the program degenerates to simple spinning. <strong>An in fact that's the litmus test for correct <code>park()</code>-<code>unpark()</code> usage.</strong></p> </blockquote> <p>Empty <code>park()</code> and <code>unpark()</code> methods do not give you any <em>happens-before</em> relationship guarantees, so for your program to be 100% portable, you should not rely on them.</p> <p>Then again, the <a href="http://docs.oracle.com/javase/7/docs/api/java/util/concurrent/locks/LockSupport.html" rel="nofollow">Javadoc of LockSupport</a> says:</p> <blockquote> <p>These methods are designed to be used as tools for creating higher-level synchronization utilities, and are not in themselves useful for most concurrency control applications. The <code>park</code> method is designed for use only in constructions of the form:</p> <p><code>while (!canProceed()) { ... LockSupport.park(this); }</code></p> </blockquote> <p>Since you have to explicitly check some condition anyway, which will either involve <code>volatile</code> or properly synchronized variables, the weak guarantees of <code>park()</code> should not actually be problem, right?</p>
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    1. COThanks for this answer. I can't now remember why I wanted to know this. I suspect I was wondering if when the running thread updated the condition and called `unpark`, if that would by itself guarantee the unparked thread would see the updated condition in a consistent state. It seems there is no guarantee, so the only safe choice is to explicitly arrange a memory barrier for the condition update.
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    2. COI see. If there were a *happens-before* relationship between `park()` and `unpark()`, then the condition would not need to involve a memory barrier. So the question is still valid.
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