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Re: Bash cannot kill itself?


From: Chris F.A. Johnson
Subject: Re: Bash cannot kill itself?
Date: Wed, 30 Jun 2010 02:42:00 -0400 (EDT)
User-agent: Alpine 2.00 (LMD 1167 2008-08-23)

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.

-- 
   Chris F.A. Johnson, <http://cfajohnson.com>
   Author:
   Pro Bash Programming: Scripting the GNU/Linux Shell (2009, Apress)
   Shell Scripting Recipes: A Problem-Solution Approach (2005, Apress)



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