Re: why are some atomic_t's not volatile, while most are?

["Robert P. J. Day" <rpjday@mindspring.com>]

On Wed, 8 Aug 2007, Chris Snook wrote:

> Jerry Jiang wrote:
> > On Wed, 08 Aug 2007 02:47:53 -0400
> > Chris Snook <csnook@redhat.com> wrote:
> >
> > > Chris Friesen wrote:
> > > > Chris Snook wrote:
> > > >
> > > > > This is not a problem, since indirect references will cause the CPU to
> > > > > fetch the data from memory/cache anyway.
> > > > Isn't Zan's sample code (that shows the problem) already using indirect
> > > > references?
> > > Yeah, I misinterpreted his conclusion.  I thought about this for a while,
> > > and realized that it's perfectly legal for the compiler to re-use a value
> > > obtained from atomic_read.  All that matters is that the read itself was
> > > atomic.  The use (or non-use) of the volatile keyword is really more
> > > relevant to the other atomic operations.  If you want to guarantee a
> > > re-read from memory, use barrier().  This, incidentally, uses volatile
> > > under the hood.
> > >
> >
> >
> > So for example, without volatile
> >
> > int a = read_atomic(v);
> > int b = read_atomic(v);
> >
> > the compiler will optimize it as b = a, But with volatile, it will be forced
> > to fetch v's value from memory
> > again.
> >
> > So, come back our initial question,
> > include/asm-v850/atomic.h:typedef struct { int counter; } atomic_t;
> >
> > Why is it right without volatile?
>
> Because atomic_t doesn't promise a memory fetch every time.  It merely
> promises that any atomic_* operations will, in fact, be atomic.  For example,
> posted today:
>
> http://lkml.org/lkml/2007/8/8/122

i'm sure that, when this is all done, i'll finally have an answer to
my original question, "why are some atomic_t's not volatile, while
most are?"

i'm almost scared to ask any more questions.  :-)

rday
-- 
========================================================================
Robert P. J. Day
Linux Consulting, Training and Annoying Kernel Pedantry
Waterloo, Ontario, CANADA

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