Re: [patch] cpufreq: mark cpufreq_tsc() as core_initcall_sync

On 11/20, Paul E. McKenney wrote:

On Mon, Nov 20, 2006 at 09:57:12PM +0300, Oleg Nesterov wrote:

So, if we have global A == B == 0,

CPU_0 CPU_1

A = 1; B = 2;
mb(); mb();
b = B; a = A;

It could happen that a == b == 0, yes? Isn't this contradicts with definition
of mb?

It can and does happen. -Which- definition of mb()? ;-)

I had a somewhat similar understanding before this discussion

[PATCH] Fix RCU race in access of nohz_cpu_mask
http://marc.theaimsgroup.com/?t=113378060600003

Semantics of smp_mb() [was : Re: [PATCH] Fix RCU race in access of nohz_cpu_mask ]
http://marc.theaimsgroup.com/?t=113432312600001

Could you please explain me again why that fix was correct? What we have now is:

CPU_0 CPU_1
rcu_start_batch: stop_hz_timer:

rcp->cur++; STORE nohz_cpu_mask |= cpu

smp_mb(); mb(); // missed actually

->cpumask = ~nohz_cpu_mask; LOAD if (rcu_pending()) // reads rcp->cur
nohz_cpu_mask &= ~cpu

So, it is possible that CPU_0 reads an empty nohz_cpu_mask and starts a grace
period with CPU_1 included in rcp->cpumask. CPU_1 in turn reads an old value
of rcp->cur (so rcu_pending() returns 0) and becomes CPU_IDLE.

Take another patch,

Re: Oops on 2.6.18
http://marc.theaimsgroup.com/?l=linux-kernel&m=116266392016286

switch_uid: __sigqueue_alloc:

STORE 'new_user' to ->user STORE "locked" to ->siglock

mb(); "mb()"; // sort of, wrt loads/stores above

LOAD ->siglock LOAD ->siglock

Agian, it is possible that switch_uid() doesn't notice that ->siglock is locked
and frees ->user. __sigqueue_alloc() in turn reads an old (freed) value of ->user
and does get_uid() on it.

To see how this can happen, thing of the SMP system as a message-passing
system, and consider the following sequence of events:

o The cache line for A is initially in CPU 1's cache, and the
cache line for B is initially in CPU 0's cache (backwards of
what you would want knowing about the upcoming writes).

o CPU 0 stores to A, but because A is not in cache, places it in
CPU 0's store queue. It also puts out a request for ownership
of the cache line containing A.

o CPU 1 stores to B, with the same situation as for CPU 0's store
to A.

o Both CPUs execute an mb(), which ensures that any subsequent writes
follow the writes to A and B, respectively. Since neither CPU
has yet received the other CPU's request for ownership, there is
no ordering effects on subsequent reads.

o CPU 0 executes "b = B", and since B is in CPU 0's cache, it loads
the current value, which is zero.

o Ditto for CPU 1 and A.

o CPUs 0 and 1 now receive each other's requests for ownership, so
exchange the cache lines containing A and B.

o Once CPUs 0 and 1 receive ownership of the respective cache lines,
they complete their writes to A and B (moving the values from the
store buffers to the cache lines).

Paul, Alan, in case it was not clear: I am not arguing, just trying to
understand, and I appreciate very much your time and your explanations.

Oleg.

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