Re: Additional compiler barrier required in sched_preempt_enable_no_resched?

[Peter Zijlstra <peterz@infradead.org>]

On Mon, May 16, 2016 at 12:55:57PM +0200, Peter Zijlstra wrote:
> You're right in that the 'proper' sequence:
> 
> > #define preempt_enable() \
> > do { \
> >         barrier(); \
> >         if (unlikely(preempt_count_dec_and_test())) \
> >                 __preempt_schedule(); \
> > } while (0)
> 
> > #define preempt_disable() \
> > do { \
> >         preempt_count_inc(); \
> >         barrier(); \
> > } while (0)
> 
> Has a higher chance of succeeding to emit the operations to memory; but
> an even smarter pants compiler might figure doing something like:
> 
> 	if (preempt_count() == 1)
> 		__preempt_schedule();
> 
> is equivalent and emits that instead, not bothering to modify the actual
> variable at all -- the program as specified cannot tell the difference
> etc..

For this to work the call __preempt_schedule() must be obfuscated
though; but I think the thing we do for x86 which turns it into:

	asm("call ___preempt_schedule;");

might just be enough for the compiler to get confused about that.

A normal function call would act as a compiler barrier and force the
compiler to emit the memory ops.

Then again, it could maybe do:

	if (preempt_count() == 1) {
		preempt_count_dec();
		__preempt_schedule();
		preempt_count_inc();
	}

and think it did us a service by 'optimizing' away that memory reference
in the 'fast' path.

Who knows what these compilers get up to ;-)