Ah, excellent! I seem to recall a post somewhere mentioning that relying on volatile might not always work on multiprocessor systems. Is that something to worry about?

Quick follow-up question:
Why do you use auto_ptr<bool> when passing the flag_guard to the thread when you release it right afterwards? Is there a chance that thread's constructor or bind() might throw an exception?

Thanks for the quick response,
Soren



> Date: Wed, 27 Dec 2006 20:45:40 +0100
> From: sebastian.redl@getdesigned.at
> To: boost-users@lists.boost.org
> Subject: Re: [Boost-users] [Boost.Threads] Exiting Thread Properly
>
> Soren Dreijer wrote:
> > I'm wondering what the best approach is to signal a
> > thread to exit.
> Just allocate a boolean variable for each thread and pass the address to
> the thread starting function. The loop of the thread breaks if the flag
> is reset; the thread exits. This scheme is so simple that it doesn't
> even require synchronization beyond making the pointer the thread
> accesses volatile.
> You could use a shared_ptr to manage the memory, but personally I would
> simply use a scoped_ptr in the thread's main function.
> The following code is conceptual. I can't remember the exact API of
> Boost.Threads.
>
> void thread_proc(volatile bool *flag)
> {
> scoped_ptr<bool> guard(flag);
> while(*flag) {
> // Work
> }
> }
>
> map<threadid, bool *> flags;
>
> void create_worker()
> {
> auto_ptr<bool> flag_guard(new bool(true));
> thread t(bind(&thread_proc, flag_guard.get()));
> // Thread is started. This ensures that it will release the memory.
> bool *b = flag_guard.release();
>
> flags.insert(make_pair(t.get_id(), b));
> }
>
> // Note: this function needs to synchronize if you can't ensure that it
> will only be ever called once
> // for any given thread id.
> void signal_worker(threadid id)
> {
> flag_iterator i = flags.find(id);
> if(i != flags.end()) {
> bool *b = i->second;
> flags.remove(id);
> *b = false;
> }
> }
>
> You will need to synchronize map access, of course.
>
> Sebastian Redl
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