Re: Control dependency between prior load in while condition and later store?

[Alan Stern <stern@rowland.harvard.edu>]

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On Thu, 5 Apr 2018, Peter Zijlstra wrote:

> On Wed, Apr 04, 2018 at 04:35:32PM -0400, Alan Stern wrote:
> > On Wed, 4 Apr 2018, Daniel Jordan wrote:
> > 
> > > A question for memory-barriers.txt aficionados.
> > > 
> > > Is there a control dependency between the prior load of 'a' and the 
> > > later store of 'c'?:
> > > 
> > >    while (READ_ONCE(a));
> > >    WRITE_ONCE(c, 1);
> > 
> > I would say that yes, there is.
> 
> Indeed.
> 
> > Yes, except that a more accurate view of the object code would be
> > something like this:
> > 
> > Loop:	r1 = READ_ONCE(a);
> > 	if (r1)
> > 		goto Loop;
> > 	else
> > 		;	// Do nothing
> > 	WRITE_ONCE(c, 1);
> > 
> > Here you can see that one path branches backward, so everything 
> > following the "if" is dependent on the READ_ONCE.
> 
> Agreed, and I think I even have code that relies on such a pattern
> somewhere.. Ah.. yes, see smp_cond_load_acquire().

One does have to be very careful when talking about compiler behavior.  
This happens to be a particularly delicate point.  My old copy of the
C++11 draft standard says (section 1.10 paragraph 24):


The implementation may assume that any thread will eventually do one of 
the following:

â- terminate,
â- make a call to a library I/O function,
â- access or modify a volatile object, or
â- perform a synchronization operation or an atomic operation.

[ Note: This is intended to allow compiler transformations such as 
removal of empty loops, even when termination cannot be proven. - end 
note ]


In this example, READ_ONCE() is in fact a volatile access, so we're 
okay.  But if it weren't, the compiler might decide to assume the loop 
will eventually terminate, meaning that the WRITE_ONCE() would always 
be executed eventually.  Then there would be nothing to prevent the 
compiler from moving the WRITE_ONCE() up before the start of the loop, 
which would of course destroy the control dependency.

Alan