linux-kernel - Re: [tip:x86/urgent] x86/tls: Don't validate lm in set_thread

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Message-ID: <CA+55aFwULSs-xiyQbpUezx3ebJOfRHz4uBE3Vw7S0dXVWbfEhQ@mail.gmail.com>
Date:	Thu, 18 Dec 2014 11:03:33 -0800
From:	Linus Torvalds <torvalds@...ux-foundation.org>
To:	"H. Peter Anvin" <hpa@...or.com>
Cc:	Linux Kernel Mailing List <linux-kernel@...r.kernel.org>,
	Andy Lutomirski <luto@...capital.net>,
	Ingo Molnar <mingo@...nel.org>,
	Thomas Gleixner <tglx@...utronix.de>,
	"linux-tip-commits@...r.kernel.org" 
	<linux-tip-commits@...r.kernel.org>
Subject: Re: [tip:x86/urgent] x86/tls: Don't validate lm in set_thread_area()
 after all

On Thu, Dec 18, 2014 at 8:59 AM, H. Peter Anvin <hpa@...or.com> wrote:
>>
>> will leave .lm uninitialized.  This means that anything in the
>> kernel that reads user_desc.lm for 32-bit tasks is unreliable.
>
> No, it won't.  However, if you initialize this dynamically field by
> field rather than as an initializer, then you are correct.

Actually, even with a full initializer, unnamed parts of a structure
(so padding bytes between things, but for bitfields also unnamed
alignment fields etc) are basically "all bets are off". They are *not*
guaranteed to be initialized to zero.

So if you have a structure like

   struct {
       unsigned int a:5;
       unsigned int b;
   } x = { .a = 0, .b = 0 };

afaik the compiler is not guaranteed to initialize the left-over bits
in the first word. Because they simply don't "exist" as far as the C
language is concerned.

On the other hand, if you do

   struct {
        unsigned int a:5, unused:27;
        unsigned int b;
   } x = { .a = 0, .b = 0 };

then the 'unused' bits are guaranteed to be initialized to zero.

(Static allocations in the BSS are obviously zeroed for other reasons,
so there are no "left-over" bits there to worry about,. So in practice
the above is only about dynamic initializers).

                         Linus
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