linux-kernel - Re: Question regarding "Control Dependencies" in memory-barriers.txt

lists.openwall.net		lists / announce owl-users owl-dev john-users john-dev passwdqc-users yescrypt popa3d-users / oss-security kernel-hardening musl sabotage tlsify passwords / crypt-dev xvendor / Bugtraq Full-Disclosure linux-kernel linux-netdev linux-ext4 linux-hardening linux-cve-announce PHC
Open Source and information security mailing list archives

Hash Suite: Windows password security audit tool. GUI, reports in PDF.

[<prev] [next>] [<thread-prev] [thread-next>] [day] [month] [year] [list]

Message-ID: <CAJhHMCBjub+_Gg_wC2EoR5OgzwT7JTHr-wzoGp_dePUBVwEG8g@mail.gmail.com>
Date:	Tue, 5 Aug 2014 08:13:54 -0400
From:	Pranith Kumar <pranith@...ech.edu>
To:	Peter Zijlstra <peterz@...radead.org>
Cc:	Paul McKenney <paulmck@...ux.vnet.ibm.com>,
	LKML <linux-kernel@...r.kernel.org>
Subject: Re: Question regarding "Control Dependencies" in memory-barriers.txt

On Tue, Aug 5, 2014 at 3:32 AM, Peter Zijlstra <peterz@...radead.org> wrote:
>>
>> 685 This transformation loses the ordering between the load from variable 'a'
>> 686 and the store to variable 'b'.  If you are relying on this ordering, you
>> 687 should do something like the following:
>> 688
>> 689         q = ACCESS_ONCE(a);
>> 690         BUILD_BUG_ON(MAX <= 1); /* Order load from a with store to b. */
>> 691         if (q % MAX) {
>> 692                 ACCESS_ONCE(b) = p;
>> 693                 do_something();
>> 694         } else {
>> 695                 ACCESS_ONCE(b) = p;
>> 696                 do_something_else();
>> 697         }
>> 698
>>
>> How is the BUILD_BUG_ON(MAX <= 1) guaranteeing the ordering w.r.t 'a'
>> and 'b'. Shouldn't it have barrier() in both the legs of the if()
>> statement like follows:
>
> The BUILD_BUG_ON guarantees that 'q % MAX' isn't a constant, and
> therefore the compiler cannot take the conditional out.
>
> And no barrier() doesn't order anything except compiler output. The
> ordering here happens by the conditional, CPUs do not do speculative
> writes, this means it has to complete the load of q to compute the
> branch before it can execute the store of b.
>
>

I don't think the write to 'b' here is speculative since it is
happening in both the legs of the if() conditional. The write to b can
be pulled out to before the conditional. Without the barrier(), isn't
the following a valid transformation of the above?

BUILD_BUG_ON(MAX <= 1); /* this will be compiled out if MAX != 1*/
q = ACCESS_ONCE(a);
ACCESS_ONCE(b) = p; / *BUG: No ordering */
if (q % MAX) {
        do_something();
} else {
        do_something_else();
}

I don't see how it is preserving the ordering.

-- 
Pranith
--
To unsubscribe from this list: send the line "unsubscribe linux-kernel" in
the body of a message to majordomo@...r.kernel.org
More majordomo info at  http://vger.kernel.org/majordomo-info.html
Please read the FAQ at  http://www.tux.org/lkml/