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: > > > > > > > > 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. 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
