Linux Process Groups – Why is Process Not Part of Expected Group?

linuxprocessprocess-groups

I'm learning about the relationship between processes, process groups (and sessions) in Linux.

I compiled the following program…

#include <iostream>
#include <ctime>
#include <unistd.h>

int main( int argc, char* argv[] )
{
  char buf[128];
  time_t now;
  struct tm* tm_now;
  while ( true )
  {
    time( &now );
    tm_now = localtime( &now );
    strftime( buf, sizeof(buf), "%a, %d %b %Y %T %z", tm_now );
    std::cout << buf << std::endl;
    sleep(5);
  }
  return 0;
}

… to a.out and ran it as a background process like so…

a.out &

This website says the following…

Every process is member of a unique process group, identified by its
process group ID. (When the process is created, it becomes a member of
the process group of its parent.) By convention, the process group ID
of a process group equals the process ID of the first member of the
process group, called the process group leader.

Per my reading, the first sentence conflicts with the in-parentheses content: is a process a member of a unique process group, or is it a member of the process group of its parent?

I tried to investigate with ps…

ps xao pid,ppid,pgid,sid,command | grep "PGID\|a.out"
  PID  PPID  PGID   SID COMMAND
24714 23890 24714 23890 ./a.out

This tells me my a.out process is pid 24714, spawned from parent pid 23890 and part of program group 24714. To begin with, I don't understand why this pgid matches the pid.

Next, I tried to investigate the parent process…

ps xao pid,ppid,pgid,sid,command | grep "PGID\|23890"
  PID  PPID  PGID   SID COMMAND
23890 11892 23890 23890 bash
24714 23890 24714 23890 ./a.out

It makes sense to me that the parent process of my a.out is bash. At first I thought "bash's pid matches its pgid – that must be because it's the process group leader. Maybe that makes sense because bash is kind of the "first thing" that got run, from which I ran my process." But that reasoning doesn't make sense because a.out's pgid also matches its own pid.

Why doesn't a.out's pgid equal bash's pgid? That's what I would have expected, from my understanding of the quote.

Can someone clarify the relationship between pids and pgids?

Best Answer

There is no conflict; a process will by default be in a unique process group which is the process group of its parent:

$ cat pg.c
#include <stdio.h>
#include <unistd.h>
int main(void)
{
    fork();
    printf("pid=%d pgid=%d\n", getpid(), getpgrp());
}
$ make pg
cc     pg.c   -o pg
$ ./pg 
pid=12495 pgid=12495
pid=12496 pgid=12495
$

The fork splits our process into parent (12495) and child (12496), and the child belongs to the unique process group of the parent (12495). bash departs from this because it issues additional system calls:

$ echo $$
12366
$

And then in another terminal we run:

$ strace -f -o blah -p 12366

And then back in the first terminal:

$ ./pg
pid=12676 pgid=12676
pid=12677 pgid=12676
$

And then we control+c the strace, and inspect the system calls:

$ egrep 'exec|pgid' blah
12366 setpgid(12676, 12676)             = 0
12676 setpgid(12676, 12676 <unfinished ...>
12676 <... setpgid resumed> )           = 0
12676 execve("./pg", ["./pg"], [/* 23 vars */]) = 0
12676 write(1, "pid=12676 pgid=12676\n", 21 <unfinished ...>
12677 write(1, "pid=12677 pgid=12676\n", 21 <unfinished ...>

bash has used the setpgid call to set the process group, thus placing our pg process into process group unrelated to that of the shell. (setsid(2) would be another way to tweak the process group, if you're hunting for system calls.)

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