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

From: Segher Boessenkool <[email-masked]>
Date: Wed, 15 Aug 2007 22:42:08 +0200

>> 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.

Yes, accesses to volatile objects are never reordered with
respect to each other.

> 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.

Of course.  It is executed by the abstract machine, so
it will be executed by the actual machine.  On the other
hand, try

	b = 0;
	if (b)
		b = atomic_read(a);

or similar.

Segher