Is there any mechanism in Shell script alike "include guard" in C++?

Question

let's see an example: in my main.sh, I'd like to source a.sh and b.sh. a.sh, however, might have already sourced b.sh. Thus it will cause the codes in b.sh executed twice. Is there any mechanism alike "include guard" in C++?

Answer 1

If you're sourcing scripts, you are usually using them to define functions and/or variables.

That means you can test whether the script has been sourced before by testing for (one of) the functions or variables it defines.

For example (in b.sh):

if [ -z "$B_SH_INCLUDED" ]
then
    B_SH_INCLUDED=yes
    ...rest of original contents of b.sh
fi

There is no other way to do it that I know of. In particular, you can't do early exits or returns because that will affect the shell sourcing the file. You don't have to use a name that is solely for the file; you could use a name that the file always has defined.

Answer 2

In bash, an early return does not affect the sourcing file, it returns to it as if the current file were a function. I prefer this method because it avoids wrapping the entire content in if...fi.

if [ -n "$_for_example" ]; then return; fi
_for_example=`date`

Answer 3

Personally I usually use

set +o nounset # same as set -u

on most of my scripts, therefore I always turn it off and back on.

#!/usr/bin/env bash

set +u
if [ -n "$PRINTF_SCRIPT_USAGE_SH" ] ; then
    set -u
    return
else
    set -u
    readonly PRINTF_SCRIPT_USAGE_SH=1
fi

If you do not prefer nounset, you can do this

[[ -n "$PRINTF_SCRIPT_USAGE_SH" ]] && return || readonly PRINTF_SCRIPT_USAGE_SH=1