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Message-ID: <45d127c4-71e9-2959-b69c-d31c46ec721e@suse.com>
Date: Thu, 11 May 2023 15:32:33 +0200
From: Juergen Gross <jgross@...e.com>
To: Martin Wilck <mwilck@...e.com>, linux-kernel@...r.kernel.org,
linux-scsi@...r.kernel.org
Cc: "James E.J. Bottomley" <jejb@...ux.ibm.com>,
"Martin K. Petersen" <martin.petersen@...cle.com>,
stable@...r.kernel.org
Subject: Re: [PATCH] scsi: Let scsi_execute_cmd() mark args->sshdr as invalid
On 11.05.23 15:23, Martin Wilck wrote:
> On Thu, 2023-05-11 at 15:17 +0200, Juergen Gross wrote:
>>>
>>> We know for certain that sizeof(*sshdr) is 8 bytes, and will most
>>> probably remain so. Thus
>>>
>>> memset(sshdr, 0, sizeof(*sshdr))
>>>
>>> would result in more efficient code.
>>
>> I fail to see why zeroing a single byte would be less efficient than
>> zeroing
>> a possibly unaligned 8-byte area.
>
> 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.
>
> 0xffffffff8177e9d0 <scsi_normalize_sense>: nopl 0x0(%rax,%rax,1) [FTRACE NOP]
> 0xffffffff8177e9d5 <scsi_normalize_sense+5>: test %rdi,%rdi
> 0xffffffff8177e9d8 <scsi_normalize_sense+8>: movq $0x0,(%rdx)
A struct with 8 "u8" fields can be unaligned.
x86_64 can do unaligned 8-byte stores.
Other architectures can't (e.g. MIPS). And 32-bit architectures might need
2 stores.
Juergen
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