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

[Sebastian Andrzej Siewior <bigeasy@linutronix.de>]

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