linux-kernel - Re: [PATCH] x86/kbuild: enable modversions for symbols exported from asm

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Message-ID: <3bb2d553-8f3c-d58b-d646-9a3609286260@suse.com>
Date:   Wed, 14 Dec 2016 11:15:07 +0100
From:   Michal Marek <mmarek@...e.com>
To:     Dodji Seketeli <dodji@...eteli.org>
Cc:     Nicholas Piggin <npiggin@...il.com>,
        Ian Campbell <ijc@...lion.org.uk>,
        Ben Hutchings <ben@...adent.org.uk>,
        Linus Torvalds <torvalds@...ux-foundation.org>,
        Adam Borowski <kilobyte@...band.pl>,
        Greg Kroah-Hartman <gregkh@...uxfoundation.org>,
        Linux Kbuild mailing list <linux-kbuild@...r.kernel.org>,
        Debian kernel maintainers <debian-kernel@...ts.debian.org>,
        "linux-arch@...r.kernel.org" <linux-arch@...r.kernel.org>,
        Arnd Bergmann <arnd@...db.de>, Ingo Molnar <mingo@...nel.org>,
        Linux Kernel Mailing List <linux-kernel@...r.kernel.org>
Subject: Re: [PATCH] x86/kbuild: enable modversions for symbols exported from
 asm

On 2016-12-14 11:02, Dodji Seketeli wrote:
> Michal Marek <mmarek@...e.com> a écrit:
> 
>>> Libabigail does a "whole binary" analysis of types.
>>>
>>> So, consider the point of use of the type 'struct s1*'.  Even if 'struct
>>> s' is just forward-declared at that point, the declaration of struct s1
>>> is "resolved" to its definition.  Even if the definition comes later in
>>> the binary.
>>
>> But there isn't any definition of struct s1 in t1.o. Does abidiff
>> "steal" the definition from the other object file? That would be
>> legitimate, I'm just curious.
> 
> If there is another translation unit in the *same* binary that defines
> struct s1, then yes, it's "stolen", as you say.
> 
> But if in the entire binary, struct s1 is just declared (not defined),
> then it'll compare equal to any struct s1 that is defined in the
> *second* binary.

That makes sense, thanks.

Michal