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

Chris Snook
2007-08-08T20:54:49Z

Jerry Jiang wrote:
> On Wed, 08 Aug 2007 02:47:53 -0400
> Chris Snook <csnook@redhat.com> wrote:
>
>> Chris Friesen wrote:
>>> Chris Snook wrote:
>>>
>>>> This is not a problem, since indirect references will cause the CPU to
>>>> fetch the data from memory/cache anyway.
>>> Isn't Zan's sample code (that shows the problem) already using indirect
>>> references?
>> Yeah, I misinterpreted his conclusion. I thought about this for a
>> while, and realized that it's perfectly legal for the compiler to re-use
>> a value obtained from atomic_read. All that matters is that the read
>> itself was atomic. The use (or non-use) of the volatile keyword is
>> really more relevant to the other atomic operations. If you want to
>> guarantee a re-read from memory, use barrier(). This, incidentally,
>> uses volatile under the hood.
>>
>
>
> So for example, without volatile
>
> int a = read_atomic(v);
> int b = read_atomic(v);
>
> the compiler will optimize it as b = a,
> But with volatile, it will be forced to fetch v's value from memory
> again.
>
> So, come back our initial question,
>
> include/asm-v850/atomic.h:typedef struct { int counter; } atomic_t;
>
> Why is it right without volatile?

Because atomic_t doesn't promise a memory fetch every time. It merely
promises that any atomic_* operations will, in fact, be atomic. For
example, posted today:

http://lkml.org/lkml/2007/8/8/122

-- Chris