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Subject: Re: [Boost-users] Threading question
From: Gottlob Frege (gottlobfrege_at_[hidden])
Date: 2013-12-13 12:51:59

On Fri, Dec 13, 2013 at 9:25 AM, Ulf Samuelsson <boost-user_at_[hidden]>wrote:
> Typical code.
> bool Server::QueueEmpty(void) {
> bool empty;
> try {
> THREAD_WRITE("Checking SendQueue", YELLOW);
> empty = true;
> while (1) {
> boost::mutex::scoped_lock lock(Queue_Lock, boost::try_to_lock);
> if (lock) {
> empty = SendQueue.empty();
> break;
> } else {
> // Didn't get lock, retry after yield
> THREAD_WRITE("Sendstart Yielding QueueEmpty", RED);
> this_thread::yield();
> THREAD_WRITE("Sendstart Returning QueueEmpty", RED);
> }
> }
> } catch (std::exception &e) {
> DebugPort.WriteLine(e.what(), RED);
> empty = true;
> }
> return empty;
> }
> Is there an obvious problem with this code?

- Asking if a multithreaded queue is empty is typically useless. Whatever
value it returns is already out of date.

- _trying_ to grab a lock, then yielding, then retrying, can be written
much shorter:

// get the lock, (which might require waiting/yielding until you can get it)
boost::mutex::scoped_lock lock(Queue_lock);
return SendQueue.empty();

> It is using Yield, which I assumed was the right way to release the CPU.
> Using sleep feels wrong, since I do not want to sleep, I want to wait for
> the mutex.
If you want to wait for the mutex, wait for the mutex. Pass it to
scoped_lock and it will wait for you.

> The first thread implements the mutex in the same way, although not in a
> subroutine.
If all your code uses mutexes similarly, well, I think you are
fundamentally using mutexes wrong, so the bug could be anywhere.
I mean, your example code above isn't technically wrong (that I can see),
but it just isn't how mutexes are meant to be used. What you are doing is
overly complicated, and complicated tends to lead to bugs. Keep it simple.

> Am I correct in assuming, that since I use scoped_lock, the mutex will
> always be released when I exit the while loop with "break;"?
yes. Even with an exception thrown, it will always release the lock.


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