On Wed, 15 Aug 2007, Segher Boessenkool wrote:
> > "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.
>
> > 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" ...)
> > 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.
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