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Date:	Tue, 10 Dec 2013 19:28:00 +0100
From:	Ingo Molnar <mingo@...nel.org>
To:	"Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>
Cc:	Jonathan Corbet <corbet@....net>, linux-kernel@...r.kernel.org,
	laijs@...fujitsu.com, dipankar@...ibm.com,
	akpm@...ux-foundation.org, mathieu.desnoyers@...icios.com,
	josh@...htriplett.org, niv@...ibm.com, tglx@...utronix.de,
	peterz@...radead.org, rostedt@...dmis.org, dhowells@...hat.com,
	edumazet@...gle.com, darren@...art.com, fweisbec@...il.com,
	sbw@....edu, Oleg Nesterov <oleg@...hat.com>,
	Rusty Russell <rusty@...tcorp.com.au>
Subject: Re: [PATCH tip/core/locking 4/4] Documentation/memory-barriers.txt:
 Document ACCESS_ONCE()


* Paul E. McKenney <paulmck@...ux.vnet.ibm.com> wrote:

> > +for barrier() that affects only the specific accesses flagged by the
> > +ACCESS_ONCE().
> > 
> > Does not seem to be obvious enough to me - does it affect accesses 
> > to the variables referenced (but still allows accesses to separate 
> > variables reordered), or does it affect compiler-ordering of all 
> > ACCESS_ONCE() instances, instructing the compiler to preserve 
> > program order?
> 
> I cover this in the bullet item about reordering memory accesses:
> 
>  (*) The compiler is within its rights to reorder memory accesses unless
>      you tell it not to.  For example, consider the following interaction
>      between process-level code and an interrupt handler:
> 
> 	void process_level(void)
> 	{
> 		msg = get_message();
> 		flag = true;
> 	}
> 
> 	void interrupt_handler(void)
> 	{
> 		if (flag)
> 			process_message(msg);
> 	}
> 
>      There is nothing to prevent the the compiler from transforming
>      process_level() to the following, in fact, this might well be a
>      win for single-threaded code:
> 
> 	void process_level(void)
> 	{
> 		flag = true;
> 		msg = get_message();
> 	}
> 
>      If the interrupt occurs between these two statement, then
>      interrupt_handler() might be passed a garbled msg.  Use ACCESS_ONCE()
>      to prevent this as follows:
> 
> 	void process_level(void)
> 	{
> 		ACCESS_ONCE(msg) = get_message();
> 		ACCESS_ONCE(flag) = true;
> 	}
> 
> 	void interrupt_handler(void)
> 	{
> 		if (ACCESS_ONCE(flag))
> 			process_message(ACCESS_ONCE(msg));
> 	}
> 
>      Note that the ACCESS_ONCE() wrappers in interrupt_handler()
>      are needed if this interrupt handler can itself be interrupted
>      by something that also accesses 'flag' and 'msg', for example,
>      a nested interrupt or an NMI.  Otherwise, ACCESS_ONCE() is not
>      needed in interrupt_handler() other than for documentation purposes.
>      (Note also that nested interrupts do not typically occur in modern
>      Linux kernels, in fact, if an interrupt handler returns with
>      interrupts enabled, you will get a WARN_ONCE() splat.)
> 
>      This effect could also be achieved using barrier(), but ACCESS_ONCE()
>      is more selective:  With ACCESS_ONCE(), the compiler need only forget
>      the contents of the indicated memory located, while with barrier()
>      the compiler must discard the value of all memory locations that
>      it has currented cached in any machine registers.
> 
> Does that cover it?

btw.:

  s/indicated memory located/
    indicated memory location

?

So, what I don't see this statement cover (and I might be dense about 
it!) is whether two ACCESS_ONCE() macros referring to different 
variables are allowed to be reordered with each other.

If the compiler reorders:

	ACCESS_ONCE(x);
	ACCESS_ONCE(y);

to:

	ACCESS_ONCE(y);
	ACCESS_ONCE(x);

then AFAICS it still meets the "compiler need only forget the contents 
of the indicated memory located" requirement that you listed, right?

[ I have a good excuse for asking this: after a long day my IQ dropped 
  by 50 points and all that! :-) ]

Thanks,

	Ingo
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