Sending signals to a subprocess

Ken Brown kbrown@cornell.edu
Wed Oct 20 12:21:00 GMT 2010


On 10/20/2010 1:09 AM, Andy Koppe wrote:
> On 20 October 2010 04:17, Ken Brown<kbrown@cornell.edu>  wrote:
>> Emacs creates a subprocess that runs an interactive bash shell.  Emacs wants
>> to get the PGID of the foreground process group associated to the tty of
>> this shell, and it does this on Linux via TIOCGPGRP (or equally well
>> tcgetpgrp).  I think it uses the file descriptor of the master of the pty
>> for this purpose.  If you (or some other programmer reading this) could give
>> me the code for setting all this up, I could play with it and try to figure
>> out why I'm seeing a difference between Linux and Cygwin here.  I just don't
>> know how to create a subprocess, give it a terminal, etc.
>
> Here's a test along those lines that does show a difference between
> Linux and Cygwin:
>
> #include<stdio.h>
> #include<pty.h>
>
> int main(void)
> {
>    int pid, fd;
>    pid = forkpty(&fd, 0, 0, 0);
>    if (!pid)
>      sleep(2);
>    else {
>      sleep(1);
>      printf("pid=%i fd=%i pgrp=%i\n", pid, fd, tcgetpgrp(fd));
>    }
> }

Thanks, Andy.  I had no idea how to do this.

> On Linux, where it requires -lutil to link, this gives:
>
> pid=13308 fd=3 tcgetpgrp(fd)=13308

I can confirm this on my Linux system.  I mention this because 
apparently tcgetpgrp isn't the same on all Linux systems.  See below.

> On Cygwin:
>
> pid=268 fd=3 tcgetpgrp(fd)=0

Corinna made tcgetpgrp return 0 instead of -1 in some circumstances (see 
http://www.cygwin.com/ml/cygwin-patches/2009-q4/msg00045.html) because 
she saw Linux doing that.  But when I run Corinna's test on my Linux 
system, I get -1 where she got 0.  So not all Linuxes agree on what 
tcgetpgrp should do.

> Neither of those looks POSIX-compliant to me, because tcgetpgrp should
> return -1 since fd 3 isn't the controlling terminal of the calling
> process, but the Linux behaviour is rather useful. Perhaps they
> decided to apply that restriction only to the slave side?

Ken

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