linux-kernel - Re: [PATCH 3/3] early_printk: Add simple serialization to early

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Message-Id: <20171004141745.GH3521@linux.vnet.ibm.com>
Date:   Wed, 4 Oct 2017 07:17:45 -0700
From:   "Paul E. McKenney" <paulmck@...ux.vnet.ibm.com>
To:     Steven Rostedt <rostedt@...dmis.org>
Cc:     Peter Zijlstra <peterz@...radead.org>, pmladek@...e.com,
        sergey.senozhatsky@...il.com, linux-kernel@...r.kernel.org,
        mingo@...nel.org, tglx@...utronix.de
Subject: Re: [PATCH 3/3] early_printk: Add simple serialization to
 early_vprintk()

On Wed, Oct 04, 2017 at 09:04:01AM -0400, Steven Rostedt wrote:
> On Wed, 4 Oct 2017 11:08:30 +0200
> Peter Zijlstra <peterz@...radead.org> wrote:
> 
> > On Tue, Oct 03, 2017 at 06:24:22PM -0400, Steven Rostedt wrote:
> > > On Thu, 28 Sep 2017 14:18:26 +0200
> > > Peter Zijlstra <peterz@...radead.org> wrote:  
> > > >  static int early_vprintk(const char *fmt, va_list args)
> > > >  {
> > > > +	int n, cpu, old;
> > > >  	char buf[512];
> > > > +
> > > > +	cpu = get_cpu();
> > > > +	/*
> > > > +	 * Test-and-Set inter-cpu spinlock with recursion.
> > > > +	 */
> > > > +	for (;;) {
> > > > +		/*
> > > > +		 * c-cas to avoid the exclusive bouncing on spin.
> > > > +		 * Depends on the memory barrier implied by cmpxchg
> > > > +		 * for ACQUIRE semantics.
> > > > +		 */
> > > > +		old = READ_ONCE(early_printk_cpu);
> > > > +		if (old == -1) {  
> > > 
> > > If old != -1 and old != cpu, is it possible that the CPU could have
> > > fetched an old value, and never try to fetch it again?  
> > 
> > What? If old != -1 and old != cpu, we'll hit the cpu_relax() and do the
> > READ_ONCE() again. The READ_ONCE() guarantees we'll do the load again,
> > as does the barrier() implied by cpu_relax().
> 
> I'm more worried about other architectures that don't have as strong of
> a cache coherency.
> 
> [ Added Paul as he knows a lot about odd architectures ]
> 
> Is there any architecture that we support that can have the following:
> 
> 	CPU0			CPU1
> 	----			----
> 			    early_printk_cpu = 1
>  for (;;)
>    old = READ_ONCE(early_printk_cpu);
>    [ old = 1 ]
> 
> 			    early_printk_cpu = -1
> 
>    [...]
>    cpu_relax();
>    old = READ_ONCE(early_printk_cpu);
> 
>    [ but the CPU uses the cache and not the memory? ]
> 
>    old = 1;

If you use READ_ONCE(), then all architectures I know of enforce
full ordering for accesses to a single variable.  (If you don't use
READ_ONCE(), then in theory Itanium can reorder reads.)  Me, I would
argue for WRITE_ONCE() as well to prevent store tearing.

It is only when you have at least two variables and at least two threads
than things start getting really "interesting".  ;-)

							Thanx, Paul

> -- Steve
> 
> 
> > 
> > > The cmpxchg memory barrier only happens when old == -1.  
> > 
> > Yeah, so?
> > 
> > > > +			old = cmpxchg(&early_printk_cpu, -1, cpu);
> > > > +			if (old == -1)
> > > > +				break;
> > > > +		}
> > > > +		/*
> > > > +		 * Allow recursion for interrupts and the like.
> > > > +		 */
> > > > +		if (old == cpu)
> > > > +			break;
> > > > +
> > > > +		cpu_relax();
> > > > +	}
> > > >  
> > > >  	n = vscnprintf(buf, sizeof(buf), fmt, args);
> > > >  	early_console->write(early_console, buf, n);
> > > >  
> > > > +	/*
> > > > +	 * Unlock -- in case @old == @cpu, this is a no-op.
> > > > +	 */
> > > > +	smp_store_release(&early_printk_cpu, old);
> > > > +	put_cpu();
> > > > +
> > > >  	return n;
> > > >  }  
>