Re: [PATCH 0/24] make atomic_read() behave consistently across all architectures

[Satyam Sharma <satyam@infradead.org>]

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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