linux-kernel - Re: [PATCH 4.19 28/37] dmaengine: dmatest: Fix iteration non-stop logic

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: <CAHp75VeM+qwh5rHL7RDdacru0jPSB9me2aTs__jdy749dTKRng@mail.gmail.com>
Date:   Tue, 5 May 2020 15:51:16 +0300
From:   Andy Shevchenko <andy.shevchenko@...il.com>
To:     Pavel Machek <pavel@...x.de>
Cc:     Greg Kroah-Hartman <gregkh@...uxfoundation.org>,
        Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
        Stable <stable@...r.kernel.org>,
        Dan Williams <dan.j.williams@...el.com>,
        Nicolas Ferre <nicolas.ferre@...rochip.com>,
        Andy Shevchenko <andriy.shevchenko@...ux.intel.com>,
        Vinod Koul <vkoul@...nel.org>
Subject: Re: [PATCH 4.19 28/37] dmaengine: dmatest: Fix iteration non-stop logic

On Tue, May 5, 2020 at 3:37 PM Pavel Machek <pavel@...x.de> wrote:
> > So, to the point, the conditional of checking the thread to be stopped being
> > first part of conjunction logic prevents to check iterations. Thus, we have to
> > always check both conditions to be able to stop after given
> > iterations.
>
> I ... don't understand. AFAICT the code is equivalent. Both && and ||
> operators permit "short" execution... but second part of expression
> has no sideeffects, so...

..

> You are changing !a & !b into !(a | b). But that's equivalent
> expression. I hate to admit, but I had to draw truth table to prove
> that.
>
> !a & !b   0 0 -> 1
>           else -> 0
>
> !(a | b)  0 0 -> 1
>            else -> 0
>
> What am I missing?

Basic stuff. Compiler doesn't consider second part of conjunction when
first one (see operator precedence) is already false, so, it means:

a & b
0   x -> false
1   0 -> false
1   1 -> true

x is not being considered at all. So, logically it's equivalent,
run-time it's not.

-- 
With Best Regards,
Andy Shevchenko