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Re: Bash cannot kill itself?


From: Pierre Gaston
Subject: Re: Bash cannot kill itself?
Date: Wed, 30 Jun 2010 09:55:42 +0300

On Wed, Jun 30, 2010 at 9:42 AM, Chris F.A. Johnson <address@hidden>wrote:

> On Wed, 30 Jun 2010, Clark J. Wang wrote:
>
> > On Wed, Jun 30, 2010 at 1:40 PM, Jan Schampera <address@hidden>
> wrote:
> >
> > > Clark J. Wang wrote:
> > >
> > >  Running a cmd in background (by &) would not create subshell. Simple
> > >> testing:
> > >>
> > >> #!/bin/bash
> > >>
> > >> function foo()
> > >> {
> > >>    echo $$
> > >> }
> > >>
> > >> echo $$
> > >> foo &
> > >>
> > >> ### END OF SCRIPT ###
> > >>
> > >> The 2 $$s output the same.
> > >>
> > >
> > > This doesn't mean that it doesn't create a subshell. It creates one,
> since
> > > it can't replace your foreground process.
> > >
> > > This makes sense.
> >
> > It just shows that $$ does what it should do, it reports the relevant PID
> of
> > > the parent ("main") shell you use.
> >
> >
> > Then what's the problem with my script in my original mail? Seems like
> Bash
> > does not handle the signal in a real-time way.
>
>    The special variable $$ refers to the current process, even if it
>   has the same numeric value as the parent script.

?!?
$$ is just a number.


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