Re: [PATCH] atomic: improve atomic_inc_unless_negative/atomic_dec_unless_positive

From: Frederic Weisbecker
Date: Tue Mar 12 2013 - 11:03:02 EST


2013/3/12 Paul E. McKenney <paulmck@xxxxxxxxxxxxxxxxxx>:
> On Tue, Mar 12, 2013 at 12:03:23PM +0800, Ming Lei wrote:
>> On Tue, Mar 12, 2013 at 11:39 AM, Paul E. McKenney
>> <paulmck@xxxxxxxxxxxxxxxxxx> wrote:
>> >
>> > Atomic operations that return a value are required to act as full memory
>> > barriers. This means that code relying on ordering provided by these
>> > atomic operations must also do ordering, either by using an explicit
>> > memory barrier or by relying on guarantees from atomic operations.
>> >
>> > For example:
>> >
>> > CPU 0 CPU 1
>> >
>> > X = 1; r1 = Z;
>> > if (atomic_inc_unless_negative(&Y) smp_mb();
>> > do_something();
>> > Z = 1; r2 = X;
>> >
>> > Assuming X and Z are initially zero, if r1==1, we are guaranteed
>> > that r2==1. However, CPU 1 needs its smp_mb() in order to pair with
>> > the barrier implicit in atomic_inc_unless_negative().
>> >
>> > Make sense?
>>
>> Yes, it does, and thanks for the explanation.
>>
>> But looks the above example is not what Frederic described:
>>
>> "the above atomic_read() might return -1 because there is no
>> guarantee it's seeing the recent update on the remote CPU."
>>
>> Even I am not sure if adding one smp_mb() around atomic_read()
>> can guarantee that too.
>
> Frederic was likely thinking of some other scenario that would be
> broken by atomic_inc_unless_negative() failing to act as a full
> memory barrier. Here is another example:
>
>
> CPU 0 CPU 1
>
> X = 1;
> if (atomic_inc_unless_negative(&Y) r1 = atomic_xchg(&Y, -1);
> r2 = X;
>
> If atomic_inc_unless_negative() acts as a full memory barrier, then
> if CPU 0 reaches the assignment from X, the results will be guaranteed
> to be 1. Otherwise, there is no guarantee.

Your scenarios show an interesting guarantee I did not think about.
But my concern was on such a situation:

CPU 0 CPU 1

atomic_set(&X, -1)
atomic_inc(&X)
atomic_add_unless_negative(&X, 5)

On the above situation, CPU 0 may still see X == -1 and thus not add
the 5. Of course all that only make sense with datas coming along.
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