linux-kernel - Re: [PATCH] scsi: Let scsi_execute_cmd() mark args->sshdr as invalid

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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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