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Subject: Re: kexec breaks with 5.4 due to memzero_explicit
To:     Arvind Sankar <nivedita@alum.mit.edu>
Cc:     linux-kernel@vger.kernel.org, Ingo Molnar <mingo@kernel.org>,
        x86@kernel.org
References: <20191007030939.GB5270@rani.riverdale.lan>
 <28f3d204-47a2-8b4f-f6a7-11d73c2d87c8@redhat.com>
 <0f083019-61e8-7ed5-dde7-99e1aa363d9c@redhat.com>
 <20191007132005.GB269842@rani.riverdale.lan>
From:   Hans de Goede <hdegoede@redhat.com>
Message-ID: <ec3b0807-efd0-12fc-670e-98cad9fcf754@redhat.com>
Date:   Mon, 7 Oct 2019 18:56:53 +0200
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Hi,

On 07-10-2019 15:20, Arvind Sankar wrote:
> On Mon, Oct 07, 2019 at 11:10:18AM +0200, Hans de Goede wrote:
>> Hi,
>>
>> On 07-10-2019 10:50, Hans de Goede wrote:
>>> Hi,
>>>
>>> On 07-10-2019 05:09, Arvind Sankar wrote:
>>>> Hi, arch/x86/purgatory/purgatory.ro has an undefined symbol
>>>> memzero_explicit. This has come from commit 906a4bb97f5d ("crypto:
>>>> sha256 - Use get/put_unaligned_be32 to get input, memzero_explicit")
>>>> according to git bisect.
>>>
>>> Hmm, it (obviously) does build for me and using kexec still also works
>>> for me.
>>>
>>> But it seems that you are right and that this should not build, weird.
>>
>> Ok, I understand now, it seems that the kernel will happily build with
>> undefined symbols in the purgatory and my kexec testing did not hit
>> the sha256 check path (*) so it did not crash. I can reproduce this befo=
re my patch:
>=20
> Yes -- this should really be fixed. purgatory build should fail if there
> are undefined symbols, in fact the Makefile apparently is trying to do
> something to catch undefined references?
>=20
> LDFLAGS_purgatory.ro :=3D -e purgatory_start -r --no-undefined -nostdlib =
-z nodefaultlib
>=20
> This doesn't seem to actually do anything though. Anyone know of a way
> to force ld to error if the resulting object would have undefined
> symbols?

I've figured out a way to get an error for the missing symbol, I will
Cc you on the patch which I will post upstream soon.

I will also write a similar patch for s390 and post that upstream
(untested) separately.

Regards,

Hans