2015-12-23 00:50:04 +00:00

240 lines
4.8 KiB
C

/*
* Copyright (c) 2015
* The President and Fellows of Harvard College.
*
* Redistribution and use in source and binary forms, with or without
* modification, are permitted provided that the following conditions
* are met:
* 1. Redistributions of source code must retain the above copyright
* notice, this list of conditions and the following disclaimer.
* 2. Redistributions in binary form must reproduce the above copyright
* notice, this list of conditions and the following disclaimer in the
* documentation and/or other materials provided with the distribution.
* 3. Neither the name of the University nor the names of its contributors
* may be used to endorse or promote products derived from this software
* without specific prior written permission.
*
* THIS SOFTWARE IS PROVIDED BY THE UNIVERSITY AND CONTRIBUTORS ``AS IS'' AND
* ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
* IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
* ARE DISCLAIMED. IN NO EVENT SHALL THE UNIVERSITY OR CONTRIBUTORS BE LIABLE
* FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
* DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
* OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
* HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
* LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
* OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
* SUCH DAMAGE.
*/
/*
* bigfork - concurrent VM test that behaves somewhat better than
* parallelvm.
*
* This test is a mixture of forktest and parallelvm: it does nested
* forks like forktest, and aimless matrix operations like parallelvm;
* the goal is to serve as a performance benchmark more than as a
* stress test (though it can be that too) and in particular to
* exhibit less timing variance than parallelvm does. The variance is
* still fairly high, but the variance of parallelvm is horrific.
*/
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <unistd.h>
#include <err.h>
#define BRANCHES 6
/*
* 6 branches gives 64 procs at the final stage, and we want this to
* use about 4M. So each proc's memory load should be about 1/16M or
* 64K. Which is 16384 ints, or four 64x64 matrixes.
*/
#define DIM 64
static int m1[DIM*DIM], m2[DIM*DIM], m3[DIM*DIM], m4[DIM*DIM];
static const int right[BRANCHES] = {
536763422,
478946723,
375722852,
369910585,
328220902,
62977821,
};
static unsigned failures;
static
void
init(void)
{
unsigned i, j;
srandom(73771);
for (i=0; i<DIM; i++) {
for (j=0; j<DIM; j++) {
m1[i*DIM+j] = random() % 11 - 5;
}
}
}
static
void
add(int *x, const int *a, const int *b)
{
unsigned i, j;
for (i=0; i<DIM; i++) {
for (j=0; j<DIM; j++) {
x[i*DIM+j] = a[i*DIM+j] + b[i*DIM+j];
}
}
}
static
void
mul(int *x, const int *a, const int *b)
{
unsigned i, j, k;
for (i=0; i<DIM; i++) {
for (j=0; j<DIM; j++) {
x[i*DIM+j] = 0;
for (k=0; k<DIM; k++) {
x[i*DIM+j] += a[i*DIM+k] * b[k*DIM+j];
}
}
}
}
static
void
scale(int *x, const int *a, int b)
{
unsigned i, j;
for (i=0; i<DIM; i++) {
for (j=0; j<DIM; j++) {
x[i*DIM+j] = a[i*DIM+j] / b;
}
}
}
static
void
grind(void)
{
/*
* compute: m2 = m1*m1, m3 = m2+m1, m4 = m3*m3, m1 = m4 / 2
*/
mul(m2, m1, m1);
add(m3, m2, m1);
mul(m4, m3, m3);
scale(m1, m4, 2);
}
static
int
trace(void)
{
unsigned i;
int val = 0;
for (i=0; i<DIM; i++) {
val += m1[i*DIM+i];
}
while (val < 0) {
val += 0x20000000;
}
return val % 0x20000000;
}
static
pid_t
dofork(void)
{
pid_t pid;
pid = fork();
if (pid < 0) {
warn("fork");
}
return pid;
}
static
void
dowait(pid_t pid)
{
int status;
if (pid == -1) {
failures++;
return;
}
if (pid == 0) {
exit(failures);
}
else {
if (waitpid(pid, &status, 0) < 0) {
warn("waitpid(%d)", pid);
}
else if (WIFSIGNALED(status)) {
warnx("pid %d: signal %d", pid, WTERMSIG(status));
}
else if (WEXITSTATUS(status) > 0) {
failures += WEXITSTATUS(status);
}
}
}
static
void
dotest(void)
{
unsigned i, me;
pid_t pids[BRANCHES];
int t;
char msg[128];
me = 0;
for (i=0; i<BRANCHES; i++) {
pids[i] = dofork();
if (pids[i] == 0) {
me += 1U<<i;
}
grind();
t = trace();
if (t == right[i]) {
snprintf(msg, sizeof(msg),
"Stage %u #%u done: %d\n", i, me, trace());
}
else {
snprintf(msg, sizeof(msg),
"Stage %u #%u FAILED: got %d, expected %d\n",
i, me, t, right[i]);
failures++;
}
(void)write(STDOUT_FILENO, msg, strlen(msg));
}
for (i=BRANCHES; i-- > 0; ) {
dowait(pids[i]);
}
if (failures > 0) {
printf("%u failures.\n", failures);
}
else {
printf("Done.\n");
}
}
int
main(void)
{
init();
dotest();
return 0;
}