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copied!<p>I think I got it... the big hint is "only idle can be done during the initialization phase."</p> <p>First of all, a restriction on mutexes is that you must unlock in the same thread that you locked in. So each <code>unlock()</code> in a thread must be paired with a <code>lock()</code>, and we must have an equal number of them (or else we would get a deadlock).</p> <p>This means that the only way we can prevent a thread from printing multiple times is to ensure that each thread ALWAYS OWNS AT LEAST ONE MUTEX. If thread B at any time released all its mutexes, and then the CPU switched to thread A, it could just run indefinitely.</p> <p>We can't do this with 2 mutexes without deadlock, but we CAN do it with 3.</p> <p>Parent thread:</p> <pre><code> bool initDone = false; lock(m1); lock(m2); spawnChild(); while (!initDone) sleep(); while (true) { print("Parent"); unlock(m1); lock(m3); unlock(m2); lock(m1); unlock(m3); lock(m2); } </code></pre> <p>Child thread:</p> <pre><code> lock(m3); initDone = true; while (true) { lock(m1); unlock(m3); lock(m2); print("Child"); unlock(m1); lock(m3); unlock(m2); } </code></pre> <p>We start with parent owning locks 1 and 2, and child owning 3. Then they take turns releasing and acquiring locks: parent gives lock 1 to child, child gives lock 3 to parent, parent gives lock 2 to child, child gives lock 1 to parent, parent gives lock 3 to child, child gives lock 2 to parent, repeat.</p> <p>An interesting problem; I bet you see the appeal of conditional variables now, as they trivialize this.</p>

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