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Subject: Re: [PATCH] scsi: Let scsi_execute_cmd() mark args->sshdr as invalid
From:   Martin Wilck <mwilck@suse.com>
To:     Juergen Gross <jgross@suse.com>, linux-kernel@vger.kernel.org,
        linux-scsi@vger.kernel.org
Cc:     "James E.J. Bottomley" <jejb@linux.ibm.com>,
        "Martin K. Petersen" <martin.petersen@oracle.com>,
        stable@vger.kernel.org
Date:   Thu, 11 May 2023 17:59:26 +0200
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On Thu, 2023-05-11 at 15:32 +0200, Juergen Gross wrote:
> On 11.05.23 15:23, Martin Wilck wrote:
> > On Thu, 2023-05-11 at 15:17 +0200, Juergen Gross wrote:
> > > >=20
> > > > We know for certain that sizeof(*sshdr) is 8 bytes, and will
> > > > most
> > > > probably remain so. Thus
> > > >=20
> > > > =A0=A0=A0=A0=A0 memset(sshdr, 0, sizeof(*sshdr))
> > > >=20
> > > > would result in more efficient code.
> > >=20
> > > I fail to see why zeroing a single byte would be less efficient
> > > than
> > > zeroing
> > > a possibly unaligned 8-byte area.
> >=20
> > I don't think it can be unaligned. gcc seems to think the same. It
> > compiles the memset(sshdr, ...) in scsi_normalize_sense() into a
> > single
> > instruction on x86_64.
> >=20
> > 0xffffffff8177e9d0 <scsi_normalize_sense>:=A0=A0=A0=A0=A0 nopl=A0=A0
> > 0x0(%rax,%rax,1) [FTRACE NOP]
> > 0xffffffff8177e9d5 <scsi_normalize_sense+5>:=A0=A0=A0 test=A0=A0 %rdi,%=
rdi
> > 0xffffffff8177e9d8 <scsi_normalize_sense+8>:=A0=A0=A0 movq=A0=A0 $0x0,(=
%rdx)
>=20
> A struct with 8 "u8" fields can be unaligned.

Right. I wrongly assumed this would be aligned like an u64. "The
alignment of any given struct or union type is required by the ISO C
standard to be at least a perfect multiple of the lowest common
multiple of the alignments of all of the members of the struct".

I wonder if this (non-)alignment of struct scsi_sense_hdr is
intentional, but that's a different discussion.

Thanks,
Martin