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    <p>Usually you use a <em>producer-consumer-queue</em> for this type of work.</p> <p>Whenever the worker thread runs out of work he <code>wait()</code>s on a <code>boost::condition_variable</code> which is protected by the same <code>boost::mutex</code> as the stack holding the data for the worker thread (you might want to use a queue here instead to minimize the risk of unfair work scheduling).</p> <p>Your <code>PushFrame()</code> function now calls <code>notify_one()</code> on that condition variable whenever it inserts new data into the stack. That way, the worker thread will truly sleep (i.e. the OS scheduler will probably not give it any timeslice) until there is actually work to be done.</p> <p>The easiest thing to get wrong here is the locking of the mutex protecting both the stack and the condition_variable. Besides avoiding races on the data structures, you also need to take care that the condition_variable does not miss a notify call and therefore might get stuck waiting while there actually is more work available.</p> <pre><code>class Worker { void PushFrame(byte* data) { boost::lock_guard&lt;boost::mutex&gt; lk(m_mutex); // push the data // ... m_data_cond.notify_one(); } void DoWork() { while(!done) { boost::unique_lock&lt;boost::mutex&gt; lk(m_mutex); // we need a loop here as wait() may return spuriously while(is_out_of_work()) { // wait() will release the mutex and suspend the thread m_data_cond.wait(lk); // upon returning from wait() the mutex will be locked again } // do work from the queue // ... } } boost::mutex m_mutex; boost::condition_variable m_data_cond; [...] }; </code></pre>
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    1. USComicSansMS
    1. USComicSansMS
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    1. COnotify_one is a part of condition_variable. Should the worker inherit from it?Can you add some simple example of the concept? Thanks!
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      1. USMichael IV
    2. COThe condition_variable should probably be a member of your Worker class. Added a small code example to the answer.
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    3. CO@ComicSansMS I think the comment `do work from the queue` should go at the end of the loop (after checking for work), otherwise the queue could be empty. This could happen on first running `DoWork()` (if no work has yet been posted) or if there are multiple Worker threads and another thread grabs the mutex after the lock is released at the end of the loop.
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