Re: [ECOS] Condition Variables recap

[Jonathan Larmour <jlarmour at redhat dot com>]

"Trenton D. Adams" wrote:
> 
>   >
>   > "Trenton D. Adams" wrote:
>   > >
>   > > Consumer Thread
>   > > Each consumer thread starts in a loop and checks the condition
>   > variable
>   > > by calling cyg_cond_wait ().
>   > > They will either be suspended, or one will succeed.
>   > > After the producer thread calls cyg_cond_broadcast (), ONE
> consumer
>   > > thread will lock the mutex, and begin reading data.
>   > > Now this thread should call cyg_mutex_unlock () which will cause
>   > another
>   > > ONE thread to lock it.
>   > >
>   > > Now I have a question.  What happens to the cyg_cond_wait () call
> in
>   > the
>   > > OTHER consumer threads when cyg_cond_broadcast () is called?  Do
> they
>   > > simply get suspended, after that other ONE consumer thread locks
> the
>   > > mutex, and until the mutex is available to ONE of them again?
>   >
>   > Yes. They will all have been woken up by the broadcast, and then
> each go
>   > through locking and unlocking the mutex. By that point the condition
> may
>   > no
>   > longer be true of course, i.e. there may be no more waiting data. I
>   > would
>   > imagine cyg_cond_signal() would be more appropriate (although
> there's
>   > still
>   > a chance more than one thread is woken up).
>   >
> 
> Why would cyg_cond_signal () be more appropriate?

Shouldn't only one thread need to process the arrived data? With
_broadcast(), all of them will wake up. But still, I don't know your
application, so it's up to you. If it works for you, whatever :).
 
> Is there any way that two of the threads could accidentally lock a mutex
> at the exact same time and not even know it?

No. Unless you misuse the API (e.g. by calling cyg_mutex_trylock() and
don't check the return code).

Jifl
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