Is passing a different $0 to a child bash script possible?

Question

I know that this may sound as "not as intended by designer" but I have real life situation where a bash script I am modifying needs to call another bash script that I am not allowed to modify.

That "unmodifiable bash script" starts with:

source `dirname $0`/setenv.sh

(that setenv.sh also starts with SCR2PATH=source "$( cd "$(dirname "$0")" ; pwd -P )" and is also unmodifiable)

Is there a trick in which I can fool the child script with a different $0 than that of the calling script?

Answer 1

Maybe this will do:

bash -c '. /path/to/unmodifiable' /fake/path args ...

Obviously, /path/to/unmodifiable should be a shell script; this silly trick won't work with other kind of executables. This also won't work with zsh (but will work with other shells like dash, ksh, etc).

If you have to source instead of call the unmodifiable script, you can use the same trick from a wrapper calling your outer script, or have it re-call itself:

$ cat unmodifiable
#! /bin/sh
printf '{%s} ' "$0" "$@"
echo

$ cat becoming
#! /bin/sh
[ "$MYSELF" ] || MYSELF=$0 exec sh -c '. "$MYSELF"' "I'm the mountain!" "$@"
. ./unmodifiable

$ chmod 755 becoming
$ ./becoming 1 2 3
{I'm the mountain!} {1} {2} {3}

Answer 2

You cannot "fool" the child script, but you can call it via a different name by creating a symlink to the called script and calling the symlink. This will, I believe achieve the objective you are looking for.

Example:

$ cat ./bin/script.sh
echo $0

$ ./bin/script.sh
./bin/script.sh
$ ln -s bin/script.sh foo.sh
$ ./foo.sh
./foo.sh

Answer 3

You can use exec -a in a subshell: (exec -a fakeDollarZero /unmodifiable/bash/script the rest of the args)

Unlike mosvy's answer, this would work even if it were something other than a bash script.