Return an exit code without closing shell

Question

I'd like to return an exit code from a BASH script that is called within another script, but could also be called directly. It roughly looks like this:

#!/bin/bash dq2-get $1 if [ $? -ne 0 ]; then   echo "ERROR: ..."   # EXIT HERE fi # extract, do some stuff # ... 

Now in the line EXIT HERE the script should exit and return exit code 1. The problem is that

• I cannot use return, because when I forget to source the script instead of calling it, return will not exit, and the rest of the script will be executed and mess things up.
• I cannot use exit, because this closes the shell.
• I cannot use the nice trick kill -SIGINT $$, because this doesn't allow to return an exit code.

Is there any viable alternative that I have overlooked?

Answer 1

The answer to the question title (not in the body as other answers have addressed) is:

Return an exit code without closing shell

(exit 33) 

If you need to have -e active and still avoid exiting the shell with a non-zero exit code, then do:

(exit 33) && true 

The true command is never executed but is used to build a compound command that is not exited by the -e shell flag.

That sets the exit code without exiting the shell (nor a sourced script).

For the more complex question of exiting (with an specific exit code) either if executed or sourced:

#!/bin/bash [ "$BASH_SOURCE" == "$0" ] &&     echo "This file is meant to be sourced, not executed" &&          exit 30  return 88 

Will set an exit code of 30 (with an error message) if executed.
And an exit code of 88 if sourced. Will exit both the execution or the sourcing without affecting the calling shell.

Answer 2

Use this instead of exit or return:

[ $PS1 ] && return || exit; 

Works whether sourced or not.