From: David Laight <David.Laight@ACULAB.COM>
To: 'Petr Mladek' <pmladek@suse.com>,
Richard Fitzgerald <rf@opensource.cirrus.com>
Cc: "rostedt@goodmis.org" <rostedt@goodmis.org>,
"sergey.senozhatsky@gmail.com" <sergey.senozhatsky@gmail.com>,
"linux-kernel@vger.kernel.org" <linux-kernel@vger.kernel.org>,
"patches@opensource.cirrus.com" <patches@opensource.cirrus.com>,
"Andy Shevchenko" <andriy.shevchenko@linux.intel.com>,
Rasmus Villemoes <linux@rasmusvillemoes.dk>
Subject: RE: [PATCH] lib: vsprintf: Avoid 32-bit truncation in vsscanf number parsing
Date: Thu, 12 Nov 2020 16:45:46 +0000 [thread overview]
Message-ID: <1fd034ae48b342429896120cefee91d9@AcuMS.aculab.com> (raw)
In-Reply-To: <20201112161746.GH20201@alley>
From: Petr Mladek
> Sent: 12 November 2020 16:18
>
> Adding other vsprintf maintainers and reviewes into CC.
>
> On Thu 2020-11-12 11:17:59, Richard Fitzgerald wrote:
> > Number conversion in vsscanf converts a whole string of digits and then
> > extracts the field width part from the converted value. The maximum run
> > of digits is limited by overflow. Conversion was using either
> > simple_strto[u]l or simple_strto[u]ll based on the 'L' qualifier. This
> > created a difference in truncation between builds where long is 32-bit
> > and builds where it is 64-bit. This especially affects parsing a run of
> > contiguous digits into separate fields - the maximum length of the run
> > is 16 digits if long is 64-bit but only 8 digits if long is 32-bits.
> > For example a conversion "%6x%6x" would convert both fields correctly if
> > long is 64-bit but not if long is 32-bit.
If %6x%6x works one might expect %6x%6x%6x to also work.
> I might be just slow today. But it took me really long time to
> understand what exactly is the problem and how it is caused.
> The description is nicely detailed but somehow cryptic.
>
> My understanding is that there is a bug when parsing numbers
> with a limited width like the above mentioned "%6x%6x".
>
> The problem is how the width is handled:
>
> 1. The width is ignored in the 1st step. The entire number
> is read using simple_strto[u]l[l] functions.
>
> 2. The width limit is achieved by dividing the result from
> the first step until it fits the width.
>
> It gives wrong result when there was an overflow in the 1st step.
> The high bits were lost even when the limited number would
> not overflow.
What happens if there are leading zeros on the hex input?
From the description I think 'something terrible happens'.
(Well, horribly unexpected anyway.)
I'd also expect strtoull() to 'eat' all the digits that exist,
not stop when the value got too large.
David
-
Registered Address Lakeside, Bramley Road, Mount Farm, Milton Keynes, MK1 1PT, UK
Registration No: 1397386 (Wales)
prev parent reply other threads:[~2020-11-12 16:45 UTC|newest]
Thread overview: 8+ messages / expand[flat|nested] mbox.gz Atom feed top
2020-11-12 11:17 Richard Fitzgerald
2020-11-12 15:35 ` Steven Rostedt
2020-11-12 15:46 ` Richard Fitzgerald
2020-11-12 17:04 ` Steven Rostedt
2020-11-13 14:00 ` Petr Mladek
2020-11-16 10:47 ` Richard Fitzgerald
2020-11-12 16:17 ` Petr Mladek
2020-11-12 16:45 ` David Laight [this message]
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