lparstat doesn't report physc correctly. it's always 0.00.

[mmatsuya]

Hi,

physc value in the output of lparstat is always 0.00. This doesn't provide any correct data.
physc is calculated as below.

void get_cpu_physc(struct sysentry *unused_se, char *buf)
{
... snip ...
physc = (new_purr - old_purr)/timebase/elapsed;
sprintf(buf, "%.2f", physc);

purr is the value in /proc/ppc64/lparcfg. old one is the value collected in the last output.
timebase is the value in /proc/cpuinfo. It was 512000000 in my case.

It seems that elapsed's value was wrong. It should be a difference (in second) from the last output timestamp. The timestamp was collected in get_time().

void get_time()
{
struct timeval t;
struct sysentry *se;

    gettimeofday(&t, 0);

    se = get_sysentry("time");
    sprintf(se->value, "%ld", t.tv_sec + t.tv_usec);

}

t.tv_sec + t.tv_usec is wrong definitely. The unit doesn't same.
If the time is returned in second here, it may cause another problem.
I created a patch, and get_time() returns the time in usec in it.
And, it can be calculated properly with changing into second in get_cpu_physc().

--- ./src/lparstat.c.original   2016-06-15 16:55:02.209992105 +0900
+++ ./src/lparstat.c.original   2016-06-15 16:56:30.398554069 +0900
@@ -65,7 +65,7 @@ void get_time()
    gettimeofday(&t, 0);

    se = get_sysentry("time");
-   sprintf(se->value, "%ld", t.tv_sec + t.tv_usec);
+   sprintf(se->value, "%ld", t.tv_sec*1000000 + t.tv_usec);
 }

 long long elapsed_time()
@@ -120,7 +120,7 @@ void get_cpu_physc(struct sysentry *unus
    new_purr = strtoll(se->value, NULL, 0);
    old_purr = strtoll(se->old_value, NULL, 0);

-   physc = (new_purr - old_purr)/timebase/elapsed;
+   physc = (new_purr - old_purr)*1000000/(timebase*elapsed);
    sprintf(buf, "%.2f", physc);
 }

Regards,
Masahiro Matsuya