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On Wed, Jun 27, 2007 at 09:45:54PM +0200, Manuel Bouyer wrote:
> Note that this can also be dependant on the drives. 
> The attached program will test the bandwidth between the drive's cache and the
> host. usage: ./tst /dev/rwd0d 10000 (increase last parameter if it completes
> too fast to give accurate results).

here is the program

-- 
Manuel Bouyer <bouyer@antioche.eu.org>
     NetBSD: 26 ans d'experience feront toujours la difference
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#include <fcntl.h>
#include <unistd.h>
#include <sys/time.h>
#include <stdio.h>
#include <stdlib.h>

int
main(int argc, char **argv)
{
	static char buf[64*1024];
	int fd, i;
	struct timeval tv0, tv1;
	int t;

	fd = open(argv[1], O_RDONLY, 0);
	if (fd < 0) {
		perror("open");
		exit(1);
	}
	if (gettimeofday(&tv0, NULL) < 0) {
		perror("gettimeofday");
		exit(1);
	}
	for (i = 0; i < atoi(argv[2]); i++) {
		if (read(fd, buf, sizeof(buf)) != sizeof(buf)) {
			perror("read");
			exit(1);
		}
		if (lseek(fd, 0, SEEK_SET) < 0) {
			perror("seek");
			exit(1);
		}
			
	}
	if (gettimeofday(&tv1, NULL) < 0) {
		perror("gettimeofday");
		exit(1);
	}
	t = (tv1.tv_sec - tv0.tv_sec) * 1000000;
	t = t + tv1.tv_usec - tv0.tv_usec;
	printf("%d us, %f MB/s\n", t,
	    ((double)64 * (double)i / 1024) / ((double)t / 1000000));
	exit(0);
}

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