Paul, thanks.
Sorry, I can't reply today, just one note...
On 11/01, Paul E. McKenney wrote:
>
> OK, so it looks to me that this code relies on synchronize_sched()
> forcing a memory barrier on each CPU executing in the kernel.
No, the patch tries to avoid this assumption, but probably I missed
something.
> 1. A task running on CPU 0 currently write-holds the lock.
>
> 2. CPU 1 is running in the kernel, executing a longer-than-average
> loop of normal instructions (no atomic instructions or memory
> barriers).
>
> 3. CPU 0 invokes percpu_up_write(), calling up_write(),
> synchronize_sched(), and finally mutex_unlock().
And my expectation was, this should be enough because ...
> 4. CPU 1 executes percpu_down_read(), which calls update_fast_ctr(),
since update_fast_ctr does preempt_disable/enable it should see all
modifications done by CPU 0.
IOW. Suppose that the writer (CPU 0) does
percpu_done_write();
STORE;
percpu_up_write();
This means
STORE;
synchronize_sched();
mutex_unlock();
Now. Do you mean that the next preempt_disable/enable can see the
result of mutex_unlock() but not STORE?
Oleg.
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