Re: [boost] [threadpool] new version v12

Subject: Re: [boost] [threadpool] new version v12
From: Anthony Williams (anthony.ajw_at_[hidden])
Date: 2008-11-03 09:14:19

"Peter Dimov" <pdimov_at_[hidden]> writes:

> Anthony Williams:
>> Fibers are still tied to a particular thread, so thread-local
>> variables and boost::this_thread::get_id() still return the same value
>> for a nested task. This means that a task that calls future::get()
>> might find that its thread-local variables have been overwritten by a
>> nested task when it resumes. It also means that any data structures
>> keyed by the thread::id may have been altered. Finally, the nested
>> task inherits all the locks of the parent, so it may deadlock if it
>> tries to lock the same mutexes (rather than just block if it is
>> running on a separate thread).
>
> No. If task A holds a mutex M and waits for task B, and task B tries
> to lock M, you have deadlock no matter which thread executes B.

That's true if the thread runs task B whilst it is waiting. If it runs
task C which is completely unrelated and task C tries to acquire the
lock then it will deadlock even if task B then completes, which would
allow task A to resume and release the lock (since task A is suspended
whilst its thread runs task C).

> For the thread locals, they usually are altered in the manner they
> should've been altered, unless a task uses thread locals instead of
> ordinary locals and calls itself. Expecting errno to persist across a
> future wait (or across an arbitrary function call) is not a reasonable
> assumption.

That depends on the thread-local variable. errno is used everywhere,
so it is not safe to assume it is unchanged. You might expect a
library specific thread-local to persist across a future wait: the
futures library knows nothing of your code, so has no reason to write
to your thread locals. Unless of course it runs another task on your
thread....

Anthony
-- 
Anthony Williams
Author of C++ Concurrency in Action | http://www.manning.com/williams
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