Re: [PATCH tip/core/locking 4/4] Documentation/memory-barriers.txt:Document ACCESS_ONCE()

From: Paul E. McKenney
Date: Tue Dec 10 2013 - 14:02:12 EST


On Tue, Dec 10, 2013 at 07:28:00PM +0100, Ingo Molnar wrote:
>
> * Paul E. McKenney <paulmck@xxxxxxxxxxxxxxxxxx> 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
>
> ?

Good catch, fixed!

> 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?

True, but if the compiler was willing to reorder ACCESS_ONCE()'s volatile
accesses, it would be really hard to write reliable device drivers. The
standard says the following:

Access to volatile objects are evaluated strictly according to
the rules of the abstract machine.

That said, compiler writers and standards wonks will argue endlessly about
exactly what that does and does not mean. :-/

I added a sentence reading:

Of course, the compiler must also respect the order in which
the ACCESS_ONCE()s occur, though the CPU of course need not do so.

To the end of that paragraph. Does that help?

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

I know that feeling! And I cannot resist replying: "Just wait until you
reach your mid-50s!" ;-)

Thanx, Paul

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