Hi,
On Mon, 2005-11-28 at 10:27 -0600, xuehai zhang wrote:
> Your questioning makes sense to me. But I am not very sure about how to
> effectively count how many
> these operations can be completed in a second. Can you give me some hint?
Here's a quick-and-dirty wrapper for timing "something" over the space
of a few seconds (default 10) and working out how fast it went.
Obviously, you want to be running this on an otherwise-idle machine, and
with CPU frequency management disabled. It's really, really dumb, but
it only uses "time()", not any subsecond time sourcing, for its work.
Cheers,
Stephen
#include <unistd.h>
#include <stdlib.h>
#include <stdio.h>
#include <sys/types.h>
#include <sys/stat.h>
#include <sys/time.h>
#define TIME_SECS 10
static inline int time_is_before(time_t a, time_t b)
{
return ((signed) (a-b)) < 0;
}
void timeme_noop(void)
{
return;
}
void timeme_dosomething(void)
{
struct stat statbuf;
stat("/", &statbuf);
}
int run_time_loop(const char *what, void (*fn)(void))
{
time_t end;
int count=0;
end = time(NULL) + TIME_SECS;
printf("Timing %s for %d seconds: ", what, TIME_SECS);
fflush(stdout);
do {
++count;
fn();
} while (time_is_before(time(NULL), end));
printf("completed %d cycles in %d seconds\n", count, TIME_SECS);
return count;
}
void wait_for_tick(void)
{
time_t end;
printf("Waiting for fresh timer tick...");
fflush(stdout);
end = time(NULL) + 1;
while (time_is_before(time(NULL), end));
printf(" done.\n");
}
int main(void)
{
int loops_noop, loops_something;
float how_long;
wait_for_tick();
loops_noop = run_time_loop("noop", timeme_noop);
loops_something = run_time_loop("something", timeme_dosomething);
how_long = 1.0 / ((float) (loops_noop - loops_something) / (TIME_SECS));
printf ("Average time for something: %f seconds (%f ns).\n",
how_long, how_long * 1000000000.0);
}
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