How can I tell whether my script was sourced (dotted in) or executed?

There seem to be two reasons why people ask this: either they're trying to detect user errors and provide a friendly message, or they're Python programmers who want to use one of Python's most idiosyncratic features in bash.

Detecting user errors

There is one school of thought that says it's a bad idea to coddle Unix users in this way, and that if a Unix user really wants to execute your script instead of sourcing it, you shouldn't second-guess him or her. Setting that aside for now, we can rephrase the question to what's really being asked:

I want to give an error message and abort, if the user runs my script, instead of sourcing it from an interactive shell.

The key here, and the reason I've rephrased the question this way, is that you can't actually determine what the user typed, but you can determine whether the code is being interpreted by an interactive shell. You do that by checking for an i in the contents of $-: 

   1 # POSIX
   2 case $- in
   3   *i*) : ;;
   4     *) echo "You should dot me in" >&2; exit 1;;
   5 esac

Or using non-POSIX syntax: 

   1 # Bash/Ksh
   2 if [[ $- != *i* ]]; then
   3   echo "You should dot me in" >&2; exit 1
   4 fi

Of course, this doesn't work for tricky cases like "I want my file to be dotted in from a non-interactive script...". For those cases, see the first paragraph of this section.

More details/tests can be found here https://gist.github.com/dualbus/6811562

Emulating Python weirdness

Python has a strange feature: some Python scripts are both a program and a library at the same time. The script detects whether it is being executed as a program or imported (is that the word??) as a library/package, and behaves differently. For example, when being imported as a library, only functions are defined. The command line arguments are not processed, and none of the functions are actually called.

Bash can do this, but it's not a natural style, and you shouldn't be doing it. That said, people will just pester us until we show them how to do this unnatural thing, so here we go.

Bash has an internal variable called FUNCNAME that can be used to detect whether a script was sourced. Normally this variable only shows the names of functions on the call stack, but source or . works a lot like a function call, so it gets put on this stack.

Now, since this is bash, nothing can ever be straightforward. This variable only exists when you're inside of an actual function. So you can't do this check in the main body of the script. You have to write a function, call it, and peek one slot down in the call stack to see where you were before the function was called. 

   1 # Bash only
   2 
   3 libfunc1() { ...; }
   4 libfunc2() { ...; }
   5 
   6 sourced() {
   7     [[ ${FUNCNAME[1]} = source ]]
   8 }
   9 
  10 sourced && return
  11 
  12 # process command line arguments, etc.

Note that the name source gets put on the stack even if the script was dotted in using ., so this check is sufficient for both cases.

If this script is executed, it will define the three functions, fail the check, and move on to the "process command line arguments, etc." section, which presumably contains some commands. If the script is sourced or dotted in, it will only define the three functions, and nothing more.

Discussion

Doesn't return 2>/dev/null work just as well to "emulate Python's weirdness"? —Crestwave

I think (( ${#BASH_SOURCE[@]} > 1 )) works better as a way to detect whether the file has been sourced. —emanuele6

Take advantage of the exit code of the $- test for more compactness: [[ $- != *[cs]* ]];(($?+${#BASH_SOURCE[@]}>1)). —pj

BashFAQ/109 (last edited 2024-08-15 17:14:29 by emanuele6)