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

Zan Lynx
2007-08-07T22:34:22Z

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On Tue, 2007-08-07 at 15:38 -0600, Chris Friesen wrote:
> Chris Snook wrote:
>
> > That's why we define atomic_read like so:
> >
> > #define atomic_read(v) ((v)->counter)
> >
> > This avoids the aliasing problem, because the compiler must de-reference
> > the pointer every time, which requires a memory fetch.
>
> Can you guarantee that the pointer dereference cannot be optimised away
> on any architecture? Without other restrictions, a suficiently
> intelligent optimiser could notice that the address of v doesn't change
> in the loop and the destination is never written within the loop, so the
> read could be hoisted out of the loop.
>
> Even now, powerpc (as an example) defines atomic_t as:
>
> typedef struct { volatile int counter; } atomic_t
>
>
> That volatile is there precisely to force the compiler to dereference it
> every single time.

I just tried this with GCC 4.2 on x86_64 because I was curious.

struct counter_t { volatile int counter; } test;
struct counter_t *tptr = &test;

int main() {
int i;

tptr->counter = 0;
i = 0;
while(tptr->counter < 100) {
i++;
}
return 0;
}

$ gcc -O3 -S t.c

a snippet of t.s:
main:
.LFB2:
movq tptr(%rip), %rdx
movl $0, (%rdx)
.p2align 4,,7
.L2:
movl (%rdx), %eax
cmpl $99, %eax
jle .L2

Now with the volatile removed:
main:
.LFB2:
movq tptr(%rip), %rax
movl $0, (%rax)
.L2:
jmp .L2

If the compiler can see it clearly, it will optimize out the load
without the volatile.
--
Zan Lynx <zlynx@acm.org>

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