Re: [PATCH] sched/task_struct: Move alloc_tag to the end of the struct.

From: Sebastian Andrzej Siewior
Date: Mon Jul 01 2024 - 04:05:48 EST


On 2024-06-30 17:23:36 [-0400], Kent Overstreet wrote:
> On Sun, Jun 30, 2024 at 11:11:42PM GMT, Sebastian Andrzej Siewior wrote:
> > On 2024-06-28 16:20:27 [-0400], Kent Overstreet wrote:
> > > > Kent, you said you didn't want it where it currently is. Fine. You said
> > > > you want it at the front next to `flags'. This isn't going to work since
> > > > there is no space left. You didn't make another suggestion or say how to
> > > > make room.
> > >
> > > It doesn't need to be on the exact same cacheline, just as near as you
> > > can get it.
> >
> > the first possible thing would be somewhere after the scheduler.
> > However, what difference does it make if it s two cache lines later or
> > more? I don't understand the requirement "closer".
>
> take advantage of CPU prefetching; CPUs will bring in more than just the
> cacheline you touched because 64 bytes is small and it's cheap to fetch
> from the same DRAM bank while it's open.

Looking at the layout:
| unsigned int flags; /* 44 4 */
| unsigned int ptrace; /* 48 4 */
| int on_cpu; /* 52 4 */
| struct __call_single_node wake_entry; /* 56 16 */
| /* --- cacheline 1 boundary (64 bytes) was 8 bytes ago --- */

Starting with sched

| struct sched_statistics stats __attribute__((__aligned__(64))); /* 704 256 */
|
| /* XXX last struct has 32 bytes of padding */
sched end, earliest spot imho

| /* --- cacheline 15 boundary (960 bytes) --- */
| unsigned int btrace_seq; /* 960 4 */

If I add this before `btrace_seq' right after `stats' then it will be 14
caches lines later or 912 bytes after. How big is this prefetch going to
be?

Sebastian