On Wed, 15 Aug 2007, Segher Boessenkool wrote:

quoted

"Volatile behaviour" itself isn't consistently defined (at least
definitely not consistently implemented in various gcc versions across
platforms),

It should be consistent across platforms; if not, file a bug please.

quoted

but it is /expected/ to mean something like: "ensure that
every such access actually goes all the way to memory, and is not
re-ordered w.r.t. to other accesses, as far as the compiler can take

                              ^
                              (volatile)

(or, alternatively, "other accesses to the same volatile object" ...)

quoted

care of these". The last "as far as compiler can take care" disclaimer
comes about due to CPUs doing their own re-ordering nowadays.

You can *expect* whatever you want, but this isn't in line with
reality at all.

volatile _does not_ prevent reordering wrt other accesses.
[...]
What volatile does are a) never optimise away a read (or write)
to the object, since the data can change in ways the compiler
cannot see; and b) never move stores to the object across a
sequence point.  This does not mean other accesses cannot be
reordered wrt the volatile access.

If the abstract machine would do an access to a volatile-
qualified object, the generated machine code will do that
access too.  But, for example, it can still be optimised
away by the compiler, if it can prove it is allowed to.

As (now) indicated above, I had meant multiple volatile accesses to
the same object, obviously.

BTW:

#define atomic_read(a)	(*(volatile int *)&(a))
#define atomic_set(a,i)	(*(volatile int *)&(a) = (i))

int a;

void func(void)
{
	int b;

	b = atomic_read(a);
	atomic_set(a, 20);
	b = atomic_read(a);
}

gives:

func:
	pushl	%ebp
	movl	a, %eax
	movl	%esp, %ebp
	movl	$20, a
	movl	a, %eax
	popl	%ebp
	ret

so the first atomic_read() wasn't optimized away.

volatile _does not_ make accesses go all the way to memory.
[...]
If you want stuff to go all the way to memory, you need some
architecture-specific flush sequence; to make a store globally
visible before another store, you need mb(); before some following
read, you need mb(); to prevent reordering you need a barrier.

Sure, which explains the "as far as the compiler can take care" bit.
Poor phrase / choice of words, probably.