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copied!<p>It looks to me like this is an attempt at optimisation gone slightly wrong. I <em>suspect</em> it's trying for "TATAS" - "Test-and-test-and-set", where it doesn't even try to do the TestAndSet if it can see the lock is already taken.</p> <p>In a <a href="http://www.bluebytesoftware.com/blog/CommentView,guid,72aaa68e-4cbd-41db-a7ce-ddc64411eafa.aspx" rel="noreferrer">post about spin locks for .NET</a>, Joe Duffy writes this TATAS code as:</p> <pre><code>class SpinLock { private volatile int m_taken; public void Enter() { while (true) { if (m_taken == 0 && Interlocked.Exchange(ref m_taken, 1) == 0) break; } } public void Exit() { m_taken = 0; } } </code></pre> <p>(Note that Joe is using 1 to mean locked and 0 to mean unlocked, unlike the code project sample - either is fine, just don't get confused between the two!)</p> <p>Note that here the call to Interlocked.Exchange <em>is conditional on <code>m_taken</code> being 0</em>. This reduces contention - the relatively expensive (I guess) test-and-set operation is avoided there it's unnecessary. I <em>suspect</em> that's what the author was aiming at, but didn't quite get it right.</p> <p>This is also mentioned in the <a href="http://en.wikipedia.org/wiki/Spin_locking" rel="noreferrer">Wikipedia article about spinlocks</a> under "significant optimisations":</p> <blockquote> <p>To reduce inter-CPU bus traffic, when the lock is not acquired, the code should loop reading without trying to write anything, until it reads a changed value. Because of MESI caching protocols, this causes the cache line for the lock to become "Shared"; then there is remarkably no bus traffic while a CPU is waiting for the lock. This optimization is effective on all CPU architectures that have a cache per CPU, because MESI is so ubiquitous.</p> </blockquote> <p>That "loop reading" is exactly what the while loop does - until it sees <code>m_taken</code> change, it only reads. When it sees the change (i.e. when the lock is released) it has another go at locking.</p> <p>Of course it's very possible that I'm missing something important - issues like this are very subtle.</p>

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