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
Attachment:
OpenPGP_0xB0DE9DD628BF132F.asc
Description: OpenPGP public key
Attachment:
OpenPGP_signature
Description: OpenPGP digital signature
