Re: [RFC][PATCH 02/11] sched: Create preempt_count invariant

[Peter Zijlstra]

On Tue, Sep 29, 2015 at 02:55:13PM +0200, Frederic Weisbecker wrote:
> On Tue, Sep 29, 2015 at 11:28:27AM +0200, Peter Zijlstra wrote:
> >  #define init_task_preempt_count(p) do { \
> > -	task_thread_info(p)->preempt_count = PREEMPT_DISABLED; \
> > +	task_thread_info(p)->preempt_count = 2*PREEMPT_DISABLED; \
> 
> Since it's not quite obvious why we use this magic value without looking
> at schedule_tail() details, maybe add a little comment? (Just "/* see schedule_tail() */").

Right, I fixed that in 12/11 v2. I'll change that around a bit.

> > +	/*
> > +	 * Still have preempt_count() == 2, from:
> > +	 *
> > +	 *	schedule()
> > +	 *	  preempt_disable();			// 1
> > +	 *	  __schedule()
> > +	 *	    raw_spin_lock_irq(&rq->lock)	// 2
> > +	 */
> 
> I found that a bit confusing first, because that's a preempt_count()
> we actually emulate for a new task. Maybe something like:
> 
> +	/*
> +	 * New task is init with preempt_count() == 2 because prev task left
> +        * us after:
> +	 *
> +	 *	schedule()
> +	 *	  preempt_disable();			// 1
> +	 *	  __schedule()
> +	 *	    raw_spin_lock_irq(&rq->lock)	// 2
> +	 */

I think I'll move the comment to finish_task_switch(), but yes.
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