On Thu, Sep 06, 2007 at 01:05:40AM +0200, Oleg Verych wrote:
quoted text
> * Sun, 2 Sep 2007 20:36:19 +0200
> >
> 
> I see, that in many places all pre-checks are done in negative form
> with resulting return or jump out. In this case, if function was called,
> what likely() path is?
> 
> > +static void resize_pid_hash(void)
> > +{
> > +	unsigned int old_shift, new_shift;
> > +
> > +	if (system_state != SYSTEM_RUNNING)
> > +		return;
> > +
> > +	old_shift = cur_pid_hash->shift;
> > +	new_shift = ilog2(nr_pids * 2 - 1);
> > +	if (new_shift == old_shift)
> > +		return;
> > +
> > +	if (!mutex_trylock(&dyn_pidhash.resize_mutex))
> > +		return;
> 
> that one or this?
> 
> ==
> 	if (system_state == SYSTEM_RUNNING) {
> 		old_shift = cur_pid_hash->shift;
> 		new_shift = ilog2(nr_pids * 2 - 1);
> 		if (new_shift != old_shift && mutex_trylock(&dyn_pidhash.resize_mutex)) {
> ==
> 		> +	old_shift = cur_pid_hash->shift;
> 		> +	new_shift = ilog2(nr_pids * 2 - 1);
> 
> /* hope this repetition is needed by design */
> 
> 		...
> 		
> 		> +	mutex_unlock(&dyn_pidhash.resize_mutex);
> 		}
> 
> What is more efficient in general sense,
> as opposed to s,3,2,1,0 Optimized?

I'm not too sure, but I'd guess that most of the time the compiler will
be able to figure out they are the same.

resize_pid_hash() fortunately isn't a fastpath anyway -- it calls
dyn_data_replace which ends up calling synchronize_rcu() 3 times,
each of which is likely to take a long time!

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