Re: open/stat64 syscalls run faster on Xen VM than standard Linux

["Stephen C. Tweedie" <sct@redhat.com>]

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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



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#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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