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rseq: Lower default slice extension

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Change the minimum slice extension to 5 usec.

Since slice_test selftest reaches a staggering ~350 nsec extension:

Task: slice_test    Mean: 350.266 ns
  Latency (us)    | Count
  ------------------------------
  EXPIRED         | 238
  0 us            | 143189
  1 us            | 167
  2 us            | 26
  3 us            | 11
  4 us            | 28
  5 us            | 31
  6 us            | 22
  7 us            | 23
  8 us            | 32
  9 us            | 16
  10 us           | 35

Lower the minimal (and default) value to 5 usecs -- which is still massive.

Signed-off-by: Peter Zijlstra (Intel) <peterz@infradead.org>
Link: https://patch.msgid.link/20260121143208.073200729@infradead.org
This commit is contained in:
Peter Zijlstra 2026-01-21 14:25:04 +01:00
parent e1d7f54900
commit 21c0e92d06
2 changed files with 2 additions and 2 deletions
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2
Documentation/userspace-api/rseq.rst
View file

@ -79,7 +79,7 @@ slice extension by setting rseq::slice_ctrl::request to 1. If the thread is
interrupted and the interrupt results in a reschedule request in the
kernel, then the kernel can grant a time slice extension and return to
userspace instead of scheduling out. The length of the extension is
determined by debugfs:rseq/slice_ext_nsec. The default value is 10 usec; which
determined by debugfs:rseq/slice_ext_nsec. The default value is 5 usec; which
is the minimum value. It can be incremented to 50 usecs, however doing so
can/will affect the minimum scheduling latency.

2
kernel/rseq.c
View file

@ -517,7 +517,7 @@ struct slice_timer {
void *cookie;
};
static const unsigned int rseq_slice_ext_nsecs_min = 10 * NSEC_PER_USEC;
static const unsigned int rseq_slice_ext_nsecs_min = 5 * NSEC_PER_USEC;
static const unsigned int rseq_slice_ext_nsecs_max = 50 * NSEC_PER_USEC;
unsigned int rseq_slice_ext_nsecs __read_mostly = rseq_slice_ext_nsecs_min;
static DEFINE_PER_CPU(struct slice_timer, slice_timer);