linux-hardening - Re: [RFC] Mitigating unexpected arithmetic overflow

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Message-ID: <CAHk-=wgXm15gZHvt8waSFhXS9yZTfvMV95xyvNtPc6FSEA_rvA@mail.gmail.com>
Date: Thu, 9 May 2024 11:08:12 -0700
From: Linus Torvalds <torvalds@...ux-foundation.org>
To: Al Viro <viro@...iv.linux.org.uk>
Cc: "Theodore Ts'o" <tytso@....edu>, Kees Cook <keescook@...omium.org>, 
	Justin Stitt <justinstitt@...gle.com>, Peter Zijlstra <peterz@...radead.org>, 
	Mark Rutland <mark.rutland@....com>, linux-hardening@...r.kernel.org, 
	linux-kernel@...r.kernel.org, llvm@...ts.linux.dev
Subject: Re: [RFC] Mitigating unexpected arithmetic overflow

On Thu, 9 May 2024 at 10:54, Al Viro <viro@...iv.linux.org.uk> wrote:
>
> On Thu, May 09, 2024 at 08:38:28AM -0700, Linus Torvalds wrote:
>
> > Going the other way is similar:
> >
> >         all_bits = low_bits + ((u64) high_bits << 16) << 16);
> >
> > and again, the compiler will recognize this idiom and do the right
> > thing (and if 'all_bits' is only 32-bit, the compiler will optimize
> > the high bit noise away).
>
> Umm...  That would make sense if it was
>         all_bits = low_bits + ((T) high_bits << 16) << 16);
> with possibly 32bit T.  But the way you wrote that (with u64) it's
> pointless - u64 _can_ be shifted by 32 just fine.

Casting to 'T' is probably a clearer option but doesn't work very well
in a helper functions that may not know what the final type is.

 Any half-way decent compiler will end up optimizing away the shifts
and adds for the high bits because they see the assignment to
'all_bits'. There's no point in generating high bits that just get
thrown away.

                Linus