Re: Bash cannot kill itself?

[Pierre Gaston]

On Wed, Jun 30, 2010 at 9:12 AM, Clark J. Wang <dearvoid@gmail.com> wrote:

> On Wed, Jun 30, 2010 at 1:40 PM, Jan Schampera <jan.schampera@web.de>
> 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 signal is delivered after the foreground process  "sleep 3600" exits,
try with:
"sleep 3600 & wait $!" instead