Re: [RFC] Strange code in cpu_idle()

[Dipankar Sarma <dipankar@in.ibm.com>]

Hello Zwane,

On Mon, Dec 06, 2004 at 02:47:11AM -0700, Zwane Mwaikambo wrote:
> Hello Paul,
> 
> On Sat, 4 Dec 2004, Paul E. McKenney wrote:
> 
> > Unless idle_cpu() is busted, it seems like the above is, given the code in
> > rcu_check_callbacks():
> > 
> > 	void rcu_check_callbacks(int cpu, int user)
> > 	{
> > 		if (user || 
> > 		    (idle_cpu(cpu) && !in_softirq() && 
> > 					hardirq_count() <= (1 << HARDIRQ_SHIFT))) {
> > 			rcu_qsctr_inc(cpu);
> > 			rcu_bh_qsctr_inc(cpu);
> > 		} else if (!in_softirq())
> > 			rcu_bh_qsctr_inc(cpu);
> > 		tasklet_schedule(&per_cpu(rcu_tasklet, cpu));
> > 	}
> > 
> > So I would say that the rcu_read_lock() in cpu_idle() is having no
> > effect, because any timer interrupt from cpu_idle() will mark a
> > quiescent state notwithstanding.  What am I missing here?
> 
> What about the hardirq_count check since we're coming in from the timer 
> interrupt?

Look at the hardirq_count check closely - it only checks for reentrant
hardirqs. If the idle task gets interrupted by a timer interrupt,
the RCU quiscent state counter for the cpu will get incremented.
So, rcu_read_lock() in cpu_idle() is bogus.

Thanks
Dipankar