linux-kernel - Re: [PATCH RFC 02/12] x86/ticketlock: convert spin loop to C

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Message-ID: <4C5C71A5.3050500@goop.org>
Date:	Fri, 06 Aug 2010 13:33:41 -0700
From:	Jeremy Fitzhardinge <jeremy@...p.org>
To:	"H. Peter Anvin" <hpa@...or.com>
CC:	Jan Beulich <JBeulich@...ell.com>,
	Peter Zijlstra <peterz@...radead.org>,
	Xen-devel <xen-devel@...ts.xensource.com>,
	Avi Kivity <avi@...hat.com>, Nick Piggin <npiggin@...e.de>,
	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>
Subject: Re: [PATCH RFC 02/12] x86/ticketlock: convert spin loop to C

  On 08/06/2010 01:17 PM, H. Peter Anvin wrote:
> On 08/06/2010 07:53 AM, Jeremy Fitzhardinge wrote:
>> On 08/06/2010 05:43 AM, Jan Beulich wrote:
>>> You certainly mean "the compiler currently treats this as being:" - I
>>> don't think there's a guarantee it'll always be doing so.
>>>
>>>> for (;;) {
>>>> if (inc.tickets.head == inc.tickets.tail)
>>>> goto out;
>>>> ...
>>>> }
>>>> out: barrier();
>>>> }
>>>>
>>>> (Which would probably be a reasonable way to clarify the code.)
>>> I therefore think it needs to be written this way.
>>
>> Agreed.
>>
>
> A call/return to an actual out-of-line function is a barrier (and will 
> always be a barrier, as it is the fundamental ABI sequence points), 
> but to an inline function it is not.

Yes. So the goto explicitly puts the barrier into the control flow which 
should stop the compiler from doing anything unexpected.

     J
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