Path: utzoo!attcan!uunet!virtech!cpcahil
From: cpcahil@virtech.uucp (Conor P. Cahill)
Newsgroups: comp.unix.questions
Subject: Re: Bourne sh/subshell question
Message-ID: <1989Nov19.221616.23167@virtech.uucp>
Date: 19 Nov 89 22:16:16 GMT
References: <1989Nov17.105828.2894@dlcq15.datlog.co.uk>
Organization: Virtual Technologies Inc.
Lines: 59

In article <1989Nov17.105828.2894@dlcq15.datlog.co.uk>, scm@dlcq15.datlog.co.uk (Steve Mawer) writes:
> (
>     echo subshell - $$
>     sleep 10
> ) &
> echo shell - $!
> 
> When run, it produces the following output:
> 
> $ sh script
> subshell - 25654
> shell - 25655
> $
> 
> My question is twofold, firstly why aren't the two PIDs identical,
> and secondly, as they're not, why is the PID of the last background
> process 1 greater than the current process PID of the subshell?

The reason for the discrepancy is that the $$ and $! are interpreted by
the parent shell (25654).  the $! (25655) is the actual process id of the
sub-shell.

Apparently the entire subshell script is processed by the shell prior to
passing it to the sub-shell.

Changing the subshell to 
	(
	    echo subshell - \$$
	    sleep 10
	) &

gets you "subshell - $$"

Using
	(
	    eval echo subshell - \$$
	    sleep 10
	) &

gets the original output.

If you need to have the correct $$ evaluation you could do something like
the following:

	sh <<\endsh &
	    echo subshell - $$
	    sleep 10
	endsh

	echo subshell = $!
	echo shell    = $$

Good luck.

-- 
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| Conor P. Cahill     uunet!virtech!cpcahil      	703-430-9247	!
| Virtual Technologies Inc.,    P. O. Box 876,   Sterling, VA 22170     |
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