Re: Bash cannot kill itself?

[Clark J. Wang]

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.

As far as I can see, this applies to all kinds of subshells like
> - explicit ones (...)
> - pipeline components
> - command substitution
> - process substitution
> - async shells (like above, running your function)
> - ...
>
> J.
>