Re: [patch 00/15] Generic Mutex Subsystem

[Balbir Singh <balbir@in.ibm.com>]

On Tue, Jan 10, 2006 at 09:03:07PM +0300, Oleg Nesterov wrote:
> Balbir Singh wrote:
> >
> > On Wed, Dec 21, 2005 at 07:42:14PM +0300, Oleg Nesterov wrote:
> > >
> > > Note that subsequent up() will not call wakeup(): ->count == 0,
> > > it just increment it. That is why we are waking the next waiter
> > > in advance. When it gets cpu, it will decrement ->count by 1,
> > > because ->sleepers == 0. If up() (++->count) was already called,
> > > it takes semaphore. If not - goes to sleep again.
> > >
> > > Or my understanding is completely broken?
> >
> > [ ... long snip ... ]
> >
> > The question now remains as to why we have the atomic_add_negative()? Why do
> > we change the count in __down(), when down() has already decremented it
> > for us?
> 
> ... and why __down() always sets ->sleepers = 0 on return.
>

I think sem->sleepers initially started as a counter, but was later converted
to a boolean value (0 or 1). The only possible values of sem->sleepers is 0, 1
or 2 and we always use sem->sleepers - 1 and set the value to either 0 or 1.

sem->sleepers is set to 0, so that when the double wakeup is called on the
wait queue, the task that wakes up (P2) corrects the count to 
(sem->sleepers - 1) = -1. This ensures that other tasks do not acquire 
the semaphore with down() (count is -1) and P2 sets sem->sleepers back to 1 
and sleeps.

 
> I don't have an answer, only a wild guess.
> 
> Note that if P1 releases this semaphore before pre-woken P2 actually
> gets cpu what happens is:
> 
> 	P1->up() just increments ->count, no wake_up() (fastpath)
> 
> 	P2 takes the semaphore without schedule.
> 
> So *may be* it was designed this way as some form of optimization,
> in this scenario P2 has some chances to run with sem held earlier.
>

P1->up() will do a wake_up() only if count < 0. For no wake_up()
the count >=0 before the increment. This means that there is no one
sleeping on the semaphore.
 
Feel free to correct me,
Balbir