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

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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 <[email protected]>

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