Re: [Boost-users] Threading question

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value it returns is already out of date.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.

Tony