How can a Bash script safely return when sourced or exit when run directly?
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MellowPine47
I have a Bash script that may either be executed directly or sourced into the current shell. When it finishes, I need it to use `return` in the sourced case and `exit` when run as a standalone script. One way to detect whether it was sourced is `(return 0 2>/dev/null)`, but that seems to encourage extra branching. I'm wondering whether this is safe: `return "$exitcode" 2>/dev/null; exit "$exitcode"`. What edge cases or unexpected behavior could this cause?
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A common approach is to put the script’s main body in a function. Then you can always use `return` inside that function, regardless of how the file was started. At the bottom, invoke the function and use its status for the standalone case, such as `main "$@"; exit $?`. This keeps the sourced-versus-executed distinction in one place instead of spreading it through the script.
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If the conditional behavior is needed at top level, you can intentionally rely on `return` failing when the file is executed directly: `return "$e" 2>/dev/null || exit "$e"`. In a sourced file, `return` succeeds and the `exit` is skipped; when executed directly, the failed `return` allows the `exit` to run.
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Be careful about where this code appears. `return` is determined by its execution context, not simply by whether the file was sourced. Inside a function, `return` is valid and succeeds whether the script was sourced or executed, so an `exit` placed afterward will never run. The fallback pattern only distinguishes the two cases when it is used at the file’s top level, outside any function.
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QuietMaple22: That can make the file structure less convenient, especially if it already contains several functions. Still, wrapping at least the main execution flow this way may be a reasonable compromise.