Also, there's no guarantee of progress. The writer can starve the reader forever.

adrian_b 26 minutes ago | parent | next [-]

That is always possible, even in any algorithm that claims to be wait-free, if some writer just keeps writing the shared data.

All the claims about something being lock-free and/or wait-free depend on a rational behavior of the writers.

If any writer acts crazy, progress becomes impossible regardless of what all others do, unless someone kills the rogue thread or process.

In practice, the algorithm from TFA is much more likely to guarantee progress than any of the algorithms that are theoretically proven to guarantee progress, because it has an extremely small overhead, while the alternatives are much more complex and they waste a lot of time.

Moreover, most wait-free algorithms guarantee progress only for the whole system, in the sense that one random thread will progress, but they do not guarantee anything for a given thread, which may be blocked forever or stuck in an infinite loop, if unlucky.

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danbruc 20 minutes ago | parent | next [-]

Not true, a wait-free algorithm guarantees that a read will complete in a bounded amount of time. And it guarantees that all threads make progress, it is lock-free that only guarantees progress for one thread.

If the value changes frequently, it will get outdated quickly, but that has nothing to do with the synchronization mechanism used. And even if writes happen rarely, there is always a chance that the value you read will be outdated a nanosecond later.

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adrian_b 15 minutes ago | parent [-]

Only the read of data small enough to be read atomically will complete in a bounded amount of time (i.e. not larger than 16 bytes on the current x86 or Arm CPUs).

If you have a bigger shared data structure, in which some other thread writes continuously, there exists absolutely no way to stop it and no way for any other thread to progress.

	▲	danbruc 6 minutes ago \| parent [-]
		No, such algorithms exist and they use various mechanisms to achieve this. For larger data structures a common trick is to make a copy, update the copy, and then replace the original or parts of it with the copy. This provide readers with a stable view of the data structure that does not depend on small atomic reads. Another mechanism is that the different threads help each other to complete their interrupted work instead of making it invalid by modifying the data right away.

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21 minutes ago | parent | prev | next [-]

[deleted]

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eqvinox 18 minutes ago | parent | prev [-]

I think you missed the point; seqlock based approaches will lock dead if you suspend/abort a thread in the wrong place. Other lock-free approaches don't have this issue. This isn't about a thread writing garbage, it's about guarantees applicable within the constraints.

	▲	adrian_b 4 minutes ago \| parent [-]
		No, you missed my point. I agree that there is the risk for a writer to be halted in the middle of its critical section, which would stop all the other writers and readers. My point is that there exists no solution that is risk free, because if a writer enters an infinite loop while writing the shared data, that will stop progress in any other algorithm, regardless if it is claimed to be wait-free. There exists no method to stop such a writer, except an external intervention from the operating system, which would have to use an IPI (inter-processor interrupt) to halt that CPU core and then kill the offending thread. In my opinion a great number of lock-free or wait-free algorithms, all of which are proposed based on the fear of what happens if a writer is halted in a critical section, are completely impractical, because their overhead is many times higher in comparison with using a lock for writers and using the method from TFA for readers. With those algorithms, a lot of CPU time is wasted continuously to guard against an event that should never happen in bug-free operating systems and applications. It is much more efficient to try to detect the lack of progress and do something about that only in the unlikely case when this happens.

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danbruc 25 minutes ago | parent | prev [-]

[...] then the readers will spin forever waiting for the counter to become even again.