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    <blockquote> <p>Am I correct in my understanding that due to out-of-order execution I can get a NullPointerException? In other words: there's no guarantee that because I read a non-null "b" I'll read a non-null "a"?</p> </blockquote> <p>Assuming that the values assigned to <code>a</code> and <code>b</code> or non-null, I think your understanding is <strong>not correct</strong>. The JLS says this:</p> <blockquote> <p>(<strong>1</strong>) If x and y are actions of the same thread and x comes before y in program order, then hb(x, y).</p> <p>(<strong>2</strong>) If an action x synchronizes-with a following action y, then we also have hb(x, y).</p> <p>(<strong>3</strong>) If hb(x, y) and hb(y, z), then hb(x, z). </p> </blockquote> <p>and</p> <blockquote> <p>(<strong>4</strong>) A write to a volatile variable (§8.3.1.4) <code>v</code> synchronizes-with all subsequent reads of <code>v</code> <em>by any thread</em> (where subsequent is defined according to the synchronization order).</p> </blockquote> <p><strong>Theorem</strong></p> <blockquote> <p>Given that thread #1 has called <code>setBoth(...);</code> once, and that the arguments were non-null, and that thread #2 has observed <code>b</code> to be non-null, then thread #2 cannot then observe <code>a</code> to be null.</p> </blockquote> <p><strong>Informal Proof</strong></p> <ol> <li>By (<strong>1</strong>) - hb(write(a, non-null), write(b, non-null)) in thread #1 </li> <li>By (<strong>2</strong>) and (<strong>4</strong>) - hb(write(b, non-null), read(b, non-null))</li> <li>By (<strong>1</strong>) - hb(read(b, non-null), read(a, XXX)) in thread #2,</li> <li>By (<strong>4</strong>) - hb(write(a, non-null), read(b, non-null))</li> <li>By (<strong>4</strong>) - hb(write(a, non-null), read(a, XXX))</li> </ol> <p>In other words, the write of a non-null value to <code>a</code> "happens-before" the read of the value (XXX) of <code>a</code>. The only way that XXX can be null, is if there was some other action writing null to <code>a</code> such that hb(write(a,non-null), write(a,XXX)) and hb(write(a,XXX), read(a,XXX)). And this is impossible according to the problem definition, and therefore XXX cannot be null. QED.</p> <p><strong>Explanation</strong> - the JLS states that the hb(...) ("happens-before") relationship does not totally forbid reordering. However, if hb(xx,yy), then reordering of actions xx and yy is <em>only</em> allowed if the resulting code has the same observable effect as the original sequence. </p>
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    1. CO@Stephen C: I should have called it *setBothNonNull* but setting first argument to null cannot happen because @NotNull is guaranteed to throw an exception (IntelliJ IDEA user since years and I started using @NotNull as soon as it was available for IntelliJ, a long time before it came to Eclipse).
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    2. CO@Stephen C: *"If x and y are actions of the same thread..."* but my question is precisely if *another* thread reads the volatiles. Reading your answer you make it sound like out-of-order execution is a non-existing concept in Java, which I don't think is true. If I were to do my non-null/null test *from the same thread where I do the assignment*, then I *know* I cannot have an NPE. My question is what if I'm doing this test from *another* thread?
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    3. CO@Stephen C: +1... I really like the very clean edit to your question. I'll just now have to grasp the nitty details of the "happens-before".
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