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Re: StaticMutex/ONCELOCK question [message #19928 is a reply to message #19927]

Tue, 03 February 2009 07:41

mirek
Messages: 14291
Registered: November 2005

Ultimate Member

Novo wrote on Tue, 03 February 2009 00:28

I couldn't understand completely several things with StaticMutex and ONCELOCK.

StaticMutex will never call destructor of a contained Mutex object. Is this meant to be?

Yes. OS will clean that up when program exits.

Quote:

#define ONCELOCK \
for(static volatile bool o_b_; !ReadWithBarrier(o_b_);) \
	for(static StaticMutex o_ss_; !o_b_;) \
		for(Mutex::Lock o_ss_lock__(o_ss_); !o_b_; BarrierWrite(o_b_, true))

How the above code actually works? Rolling Eyes

TIA

Do not get fooled by 3 'for' loops - these are just syntactic sugar to make ONCELOCK work on C statements and blocks - they in fact simulate the outer block

{
   static volatile bool o_b_;
   if(!ReadWithBarrier(o_b_)) {
       static StaticMutex mutex;
       mutex.Enter();
       {
           do_the_initialization - the statement 'body'
           BarrierWrite(o_b_);
       }
   }
}

The purpose is to avoid locking mutex in subsequent passes of ONCELOCK - you need the barrier code to do that.

Note that both compilers we use optimize the for loops away.

Mirek

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[Message index]

		StaticMutex/ONCELOCK question By: Novo on Tue, 03 February 2009 06:28
		Re: StaticMutex/ONCELOCK question By: mirek on Tue, 03 February 2009 07:41
		Re: StaticMutex/ONCELOCK question By: Novo on Tue, 03 February 2009 20:28
		Re: StaticMutex/ONCELOCK question By: Novo on Sun, 08 February 2009 03:09

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