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    <p>You could probably implement a limited-size queue with least difficulty... I was thinking about it lately and came up with this design, but you can probably find many other interesting ideas: (WARNING: it might have some issues!)</p> <ul> <li>the queue is an array of pointers to items</li> <li>you have to manage 2 pointers (head, tail) which work over the queue in the same way as a circular buffer</li> <li>if <code>head</code> == <code>tail</code>, there are no items</li> <li>if you want to <code>enqueue(ptr)</code>, Interlocked-Swap <code>tail</code> with NULL (<code>prev_tail</code> is the swapped value) <ul> <li>if <code>prev_tail == NULL</code>, try again</li> <li>if <code>prev_tail + 1</code> (with wraparound) == <code>head</code>, your queue is full</li> <li>otherwise put your <code>ptr</code> in <code>*prev_tail</code> and assign <code>prev_tail+1</code> to <code>tail</code> (watch out for buffer wrap-around)</li> </ul></li> <li>for <code>dequeue()</code> make a copy tmp_head=head and check <code>tmp_head == tail</code> <ul> <li>if it's true, return because the queue is empty</li> <li>if it's false <ul> <li>save <code>*tmp_head</code> as <code>ptr</code></li> <li>do a CAS: compare <code>head</code> with <code>tmp_head</code> swap <code>head</code> with <code>head+1</code></li> <li>if CAS failed -- start the whole function over</li> <li>if it succeeded -- return <code>ptr</code></li> </ul></li> </ul></li> </ul> <p>You can wait on both head and tail CAS operations, but if the queue is not contended, you should succeed the first time, without unnecessary locks.</p> <p>Unlimited-size queues are "a bit" harder ;) But you should be able to create a big-enough queue for most needs.</p>
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    1. POHow do I build a lockless queue?
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    1. POHow do I build a lockless queue?
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    1. COHow are you going to check if `head == tail` on dequeue without a lock?
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    2. COOk - I didn't describe that step properly. You have to loop to the beginning, instead of around the CAS itself. CAS will detect if the head was moved.
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    3. COSounds good, I agree with the expanded explanation. Just checking :).
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