On Thu, Jul 09, 2015 at 01:36:23PM -0400, Steve Litt wrote:
> By the way, I have no personal knowledge of how many actor sockets a
> listener socket can spawn off, but if I had to guess, I'd imagine 50
> would be way too low a number, if for no other reason than none of my
> current and former ISPs would have been able to serve httpd to the
> masses if 50 was the limit.
>
> Hmmm, as far as just plain processes, maybe I'll make a fork bomb and
> see how many there are before the system bogs down. That should be
> interesting.
Here's a trivial forkbomb-ish program that I've used for some experiments.
(So as to recover safely, when fork() finally fails it will sleep briefly
and exit.)
It's probably not representative of even a trivial server, but on my
1GB N270 netbook running Alpine Linux, with ~100 programs running
(ls -d /proc/[0-9]*|wc -l = 103), it reports 7880 forks.
Thanks,
Isaac Dunham
/* Written by Isaac Dunham in the year of our Lord 2015
* No rights reserved, all warranties disclaimed.
* (You can do whatever you want, but entirely at your own risk.)
*
* A small forkbomb-ish test program intended to test how many trivial
* socket-using programs can run on a processor (and their effects on
* the scheduler).
* It gets a pair of connected sockets and repeatedly forks, with
* the main process reading from the socket and the children
* writing to it.
* On resource exhaustion, all processes will sleep for 20 seconds
* and exit, allowing you to briefly observe practical effects
* before the system recovers.
*
* Compile with -DUSLEEP if you would rather see effects of a *lot* of
* syscalls. (For example, what does the scheduler do?)
*/
#include <sys/types.h>
#include <sys/socket.h>
#include <stdio.h>
#include <unistd.h>
int main(int argc, char *argv)
{
pid_t pid;
int count, sv[2];
char got;
if (socketpair(AF_UNIX, SOCK_STREAM, 0, sv))
return 32;
while ((pid = fork()) > 0) {
count += (read(sv[1], &got, 1) > 0 )? 1: 0 ;
printf("pid: %lld forks: %d\n", (long long)pid, count);
}
if (!pid) {
write(sv[0], "a", 1);
}
#ifdef USLEEP
count = 0;
while (count < 20000) {
usleep(1000);
count ++;
}
#else
sleep(20);
#endif
return 0;
}
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