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Subject: Re: Control dependency between prior load in while condition and
 later store?
To:     Alan Stern <stern@rowland.harvard.edu>
Cc:     parri.andrea@gmail.com, will.deacon@arm.com, peterz@infradead.org,
        boqun.feng@gmail.com, npiggin@gmail.com, dhowells@redhat.com,
        j.alglave@ucl.ac.uk, luc.maranget@inria.fr,
        paulmck@linux.vnet.ibm.com, akiyks@gmail.com,
        linux-kernel@vger.kernel.org,
        Steven Sistare <steven.sistare@oracle.com>,
        Pasha Tatashin <pasha.tatashin@oracle.com>
References: <Pine.LNX.4.44L0.1804041623530.1401-100000@iolanthe.rowland.org>
From:   Daniel Jordan <daniel.m.jordan@oracle.com>
Organization: Oracle
Message-ID: <4fa47ea8-e208-16d6-3b78-747049e3ee53@oracle.com>
Date:   Wed, 4 Apr 2018 17:10:05 -0400
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On 04/04/2018 04:35 PM, 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.
> 
>> I have my doubts because memory-barriers.txt doesn't talk much about
>> loops and because of what that document says here:
>>
>>     In addition, control dependencies apply only to the then-clause and
>>     else-clause of the if-statement in question.  In particular, it does
>>     not necessarily apply to code following the if-statement:
>>
>>     	q = READ_ONCE(a);
>>     	if (q) {
>>     		WRITE_ONCE(b, 1);
>>     	} else {
>>     		WRITE_ONCE(b, 2);
>>     	}
>>     	WRITE_ONCE(c, 1);  /* BUG: No ordering against the read from 'a'. */
> 
> This refers to situations where the two code paths meet up at the end
> of the "if" statement.  If they don't meet up (because one of the paths
> branches away -- especially if it branches backward) then the
> disclaimer doesn't apply, and everything following the "if" is
> dependent.

Ok, that's the part I wasn't getting: this is how the while loop changes 
the situation.

> The reason is because the compiler knows that code following the "if"
> statement will be executed unconditionally if the paths meet up, so it
> can move that code back before the "if" (provided nothing else prevents
> such motion).  But if the paths don't meet up, the compiler can't
> perform the code motion -- if it did then the program might end up
> executing something that should not have been executed!
> 
>> It's not obvious to me how the then-clause/else-clause idea maps onto
>> loops, but if we think of the example at the top like this...
>>
>>     while (1) {
>>         if (!READ_ONCE(a)) {
>>             WRITE_ONCE(c, 1);
>>             break;
>>         }
>>     }
>>
>> ...then the dependent store is within the then-clause.  Viewed this way,
>> it seems there would be a control dependency between a and c.
>>
>> Is that right?
> 
> 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.

That clears it up, thanks very much!