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

[Oleg Nesterov]

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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