11 May
2023
11 May
'23
1:23 p.m.
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)
Martin