exec vs. source spawning piped commands

[Pete Nelson]

$ bash --version
GNU bash, version 4.1.5(1)-release (x86_64-pc-linux-gnu)
Copyright (C) 2009 Free Software Foundation, Inc.
License GPLv3+: GNU GPL version 3 or later <http://gnu.org/licenses/gpl.html
>

This is free software; you are free to change and redistribute it.
There is NO WARRANTY, to the extent permitted by law.
$ uname -a
Linux host02 2.6.32-30-server #59-Ubuntu SMP Tue Mar 1 22:46:09 UTC 2011
x86_64 GNU/Linux


I'm confused about what exec is doing in this case.

Sample script t:

#!/bin/bash
echo $0 pid: $$ ppid: $PPID args: $* 1>&2
if [ -z "$1" ]; then
  exec $0 first | $0 second
  echo should not reach here
fi

output:

$ ./t
./t pid: 13746 ppid: 12823 args:
/home/user/t pid: 13747 ppid: 13746 args: first
./t pid: 13748 ppid: 13746 args: second
should not reach here

Using exec, I would expect the second line (arg: first) to have the same pid
and ppid, but it forks a new process.  All it appears exec does in this case
is expand the relative path to a full path.  Running it without exec returns
the same pattern of pid and ppid values:

$ ./t
./t pid: 13766 ppid: 12823 args:
./t pid: 13767 ppid: 13766 args: first
./t pid: 13768 ppid: 13766 args: second
should not reach here


Using source instead of exec does run the first piped command in the current
shell process, as expected.  In that case, I would have to replace the last
echo line with an exit to meet my intentions:

#!/bin/bash
echo $0 pid: $$ ppid: $PPID args: $* 1>&2
if [ -z "$1" ]; then
  source $0 first | $0 second
  exit
fi

returns:

$ ./t
./t pid: 13792 ppid: 12823 args:
./t pid: 13792 ppid: 12823 args: first
./t pid: 13794 ppid: 13792 args: second


My intention is to give the second piped process the ability to send signals
to the first using its $PPID variable.