linux-kernel - Re: [PATCH v10 6/6] x86/split_lock: Enable split lock detection by kernel parameter

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Message-ID: <CALCETrWvq9XMTse=9JTtnVYY+U2pGcF-nP=YHvEFq4htxbfGwA@mail.gmail.com>
Date:   Mon, 16 Dec 2019 10:06:03 -0800
From:   Andy Lutomirski <luto@...nel.org>
To:     David Laight <David.Laight@...lab.com>
Cc:     Andy Lutomirski <luto@...nel.org>,
        Peter Zijlstra <peterz@...radead.org>,
        "Luck, Tony" <tony.luck@...el.com>,
        "Yu, Fenghua" <fenghua.yu@...el.com>,
        Ingo Molnar <mingo@...nel.org>,
        Thomas Gleixner <tglx@...utronix.de>,
        Ingo Molnar <mingo@...hat.com>, Borislav Petkov <bp@...en8.de>,
        H Peter Anvin <hpa@...or.com>,
        "Raj, Ashok" <ashok.raj@...el.com>,
        "Shankar, Ravi V" <ravi.v.shankar@...el.com>,
        linux-kernel <linux-kernel@...r.kernel.org>,
        x86 <x86@...nel.org>, Will Deacon <will@...nel.org>
Subject: Re: [PATCH v10 6/6] x86/split_lock: Enable split lock detection by
 kernel parameter

On Mon, Dec 16, 2019 at 9:45 AM David Laight <David.Laight@...lab.com> wrote:
>
> From: Andy Lutomirski
> > Sent: 16 December 2019 17:23
> ...
> > I'm talking specifically about x86 here, where, for example, "Reads
> > are not reordered with other reads".  So READ_ONCE *does* have
> > sequencing requirements on the CPUs.
> >
> > Feel free to replace READ_ONCE with MOV in your head if you like :)
>
> I got a little confused because I thought your reference to READ_ONCE()
> was relevant.
>
> Sometimes remembering all this gets hard.
> The docs about the effects of LFENCE and MFENCE don't really help
> (they make my brain hurt).
> I'm pretty sure I've decided in the past they are almost never needed.
>

Me too.

This whole discussion is about the fact that PeterZ is sceptical that
actual x86 CPUs have as strong a memory model as the SDM suggests, and
I'm trying to understand the exact concern.  This may or may not be
directly relevant to the kernel. :)

> Usually the ordering of reads doesn't help you.
> IIRC If locations 'a' and 'b' get changed from 0 to 1 it is perfectly possible
> for one cpu to see a==0, b==1 and another a==1, b==0 even
> though both read a then b.
> (On non-alpha this may require different cpus update a and b.)
>

x86 mostly prevents this.