On 01/27/2016 10:09 PM, Hugh Dickins wrote:
On Tue, 26 Jan 2016, Felix von Leitner wrote:
Dear Linux kernel devs,
I talked to someone who uses large Linux based hardware to run a
process with huge memory requirements (think 4 GB), and he told me that
if they do a fork() syscall on that process, the whole system comes to
standstill. And not just for a second or two. He said they measured a 45
minute (!) delay before the system became responsive again.
I'm sorry, I meant 4 TB not 4 GB.
I'm not used to working with that kind of memory sizes.
Their working theory is that all the pages need to be marked copy-on-write
in both processes, and if you touch one page, a copy needs to be made,
and than just takes a while if you have a billion pages.
I was wondering if there is any advice for such situations from the
memory management people on this list.
In this case the fork was for an execve afterwards, but I was going to
recommend fork to them for something else that can not be tricked around
with vfork.
Can anyone comment on whether the 45 minute number sounds like it could
be real? When I heard it, I was flabberghasted. But the other person
swore it was real. Can a fork cause this much of a delay? Is there a way
to work around it?
I was going to recommend the fork to create a boundary between the
processes, so that you can recover from memory corruption in one
process. In fact, after the fork I would want to munmap almost all of
the shared pages anyway, but there is no way to tell fork that.
You might find madvise(addr, length, MADV_DONTFORK) helpful:
that tells fork not to duplicate the given range in the child.
Hugh
I dont know exactly what program they are running but we test RHEL with
up to 24TB
of memory and have not seen this problem. I have mmap()'d 12TB of
memory into a
parent process private, touched every page then forked a child which
wrote to every
page thereby incurring tons of ZFOD and COW faults. It takes a while to
process the
6 billion faults but the system didnt come to a halt. The time I do see
significant pauses
is when we overcommit RAM and swap space and get into an OOMkill storm.
Attached is the program:
Thanks,
Felix
PS: Please put me on Cc if you reply, I'm not subscribed to this mailing
list.
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#include <stdlib.h>
#include <unistd.h>
#include <sys/mman.h>
#include <errno.h>
#include <stdio.h>
main(int argc,char *argv[])
{
unsigned long siz, procs, itterations, cow;
char *ptr1;
char *i;
int pid, j, k, status;
if ((argc <= 1)||(argc >4)) {
printf("bad args, usage: forkoff <memsize-in-GB> #children #itterations cow:0|1\n");
exit(-1);
}
siz = ((long)atol(argv[1])*1024*1024*1024);
procs = atol(argv[2]);
itterations = atol(argv[3]);
cow = atol(argv[4]);
printf("mmaping %ld anonymous bytes\n", siz);
ptr1 = (char *)mmap((void *)0,siz,PROT_READ|PROT_WRITE,MAP_ANONYMOUS|MAP_PRIVATE,-1,0);
if (ptr1 == (char *)-1) {
printf("ptr1 = %lx\n", ptr1);
perror("");
}
if (cow) {
printf("priming parent for child COW faults\n");
// This will cause the ZFOD faults in the parent & COW faults in the children.
for (i=ptr1; i<ptr1+siz-1; i+=4096)
*i=(char)'i';
}
printf("forking %ld processes\n", procs);
k = procs;
do{
pid = fork();
if (pid == -1) {
printf("fork failure\n");
exit(-1);
} else if (!pid) {
printf("PID %d touching %d pages\n", getpid(), siz/4096);
// This will ZFOD fault if the parent didnt otherwise it will COW fault.
for (j=0; j<itterations; j++) {
for (i=ptr1; i<ptr1+siz-1; i+=4096) {
*i=(char)'i';
}
}
printf("All done, exiting\n");
exit(0);
}
} while(--k);
while (procs-- && wait(&status));
}