On Fri, 15 Dec 2000, Kees Vonk 7249 24549 wrote:
> > But you don't need to call setsid() when you fork. Why
> > looking for complicated workaround when you can do it
> > properly without any workaround. Have you ever seen an
> > example of fork that uses setsid?
>
> Ok,
>
> here is my confusion: I call the the long running process
> from the modperl script with a system() call that was why I
> was using the setsid() (in the longrunning process).
>
> I have now changed the code my long running process to:
>
> my($nOrgPID) = fork;
> exit if $nOrgPID;
> die "Could not fork: $!" unless defined $nOrgPID;
>
> close STDIN;
> close STDOUT;
> close STDERR;
>
>
> and that does not work, I have also tried putting it in the
> modperl script instead but that doesn't work either.
Kees, if you are doing system() call just put setsid back, and add this:
tie *OUT, 'Apache';
close OUT;
Apache keeps the socket tied. Tell me whether this works for you.
I'll document these things shortly and post for your review.
>
> What am I doing wrong and why would the setsid() be a
> showstopper, especially as it is called after the close
> statements (it doesn't reopen the filehandles does it).
> I could understand that it would possibly be unnecessary, but
> how could it stop this code from working?
>
> Another (more detailed) run through.
>
> 1. modperl script starts long running process (lrp) with a
> system() call.
> 2. lrp runs the code above (fork + close (tried this with
> and without setsid)).
> 3. stop apache
> 4. start apache - fails (port in use)
> 5. netstat -na reveals port in LISTEN state.
> 6. kill lrp
> 7. netstat -na reveals port no longer in use.
> 8. start apache - now starts fine
>
>
> I think I am doing something wrong somewhere, but I am not
> sure where. Is it right that I have to do the system call and
> the fork?
>
> Kees
>
>
>
_____________________________________________________________________
Stas Bekman JAm_pH -- Just Another mod_perl Hacker
http://stason.org/ mod_perl Guide http://perl.apache.org/guide
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