Re: [PATCH 8/9] per-cgroup boot time

From: Glauber Costa
Date: Tue Sep 20 2011 - 08:38:26 EST


On 09/20/2011 06:25 AM, Peter Zijlstra wrote:
On Wed, 2011-09-14 at 17:04 -0300, Glauber Costa wrote:
Record the time in which the cgroup was created. This can be
used to provide a more accurate boottime information in
cpuacct.proc.stat.

Signed-off-by: Glauber Costa<glommer@xxxxxxxxxxxxx>
---
kernel/sched.c | 8 +++++++-
1 files changed, 7 insertions(+), 1 deletions(-)

diff --git a/kernel/sched.c b/kernel/sched.c
index 4611c54..8f254d0 100644
--- a/kernel/sched.c
+++ b/kernel/sched.c
@@ -9120,6 +9120,7 @@ struct cpuacct {
/* cpuusage holds pointer to a u64-type object on every cpu */
u64 __percpu *cpuusage;
u64 __percpu *cpustat;
+ struct timespec start_time;
};

struct cgroup_subsys cpuacct_subsys;
@@ -9172,6 +9173,8 @@ static struct cgroup_subsys_state *cpuacct_create(
acct[CPUACCT_STAT_IDLE] = kstat_cpu(i).cpustat.idle;
acct[CPUACCT_STAT_IOWAIT] = kstat_cpu(i).cpustat.iowait;
}
+
+ get_monotonic_boottime(&ca->start_time);
return&ca->css;

out_free_usage:
@@ -9316,13 +9319,16 @@ static int cpuacct_proc_stat(struct cgroup *cgrp, struct cftype *cft,
unsigned int per_softirq_sums[NR_SOFTIRQS] = {0};
struct timespec boottime;
struct cpuacct *ca = cgroup_ca(cgrp);
+ struct timespec ts;
u64 *cpustat;

user = nice = system = idle = iowait =
irq = softirq = steal = cputime64_zero;
guest = guest_nice = cputime64_zero;
getboottime(&boottime);
- jif = boottime.tv_sec;
+ getboottime(&boottime);
+ ts = timespec_add(boottime, ca->start_time);
+ jif = ts.tv_sec;

for_each_possible_cpu(i) {
cpustat = per_cpu_ptr(ca->cpustat, i);


I'm confused, what does it do? You take a boot time timestamp at cgroup
creation, add that to all boot-time readings and print the result. How
does that make sense? Subtracting the start_time, maybe, that would make
the cgroup creation time 0, adding, not so much.

Boot time represent at which times the machine was booted. In this context, at which time the container/cgroup was created. So it have to be an addition.... don't really understand your question
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