In a sourced shell script set $? to a variable's value and then unset variable at same time

Question

I have a script that can be sourced in bash or ksh or it runs in bash if executed. The script stores a desired return code in ${_rc}. I can return $_rc but that leaves _rc defined in the tty. I can unset _rc afterward but that overrides the return code. If the script is not sourced I need to use exit instead of return or i get an error message

I have the following code at the end of my script

cleanup_script # unsets all variables except _rc and _was_sourced
if [ "${_was_sourced}" == 1 ];
    return ${_rc}
else
    exit ${_rc}
fi

If I run this command

typeset > before.txt
.  ./myscript
echo $?
typeset > after.txt
diff before.txt after.txt

I get (in addition to variables and functions defined by the script that I want available to the terminal)

226a236
> _rc=0
232a243
> _was_sourced=1

Which I desire to not have if possible

You can unset _was_sourced before return, but I don't think there's any way to get rid of _rc. — Barmar
– Barmar, Commented Jun 11 at 15:41
Eh, it's doable. Move the unset into a function that returns the original $? value. Problem, though, is that then you have the function escaping into the outer shell... but maybe that's considered less messy? — Charles Duffy
– Charles Duffy, Commented Jun 11 at 15:46
If there weren't a cross-shell compatibility requirement I'd think about using traps for the purpose. — Charles Duffy
– Charles Duffy, Commented Jun 11 at 15:47
What I would consider here, if it's really just TTY use at hand, is changing the variable name from _rc to _ -- that way it's a variable that'll be overwritten by the interactive shell's regular operation. — Charles Duffy
– Charles Duffy, Commented Jun 11 at 16:01
@CharlesDuffy That a good idea that works for bash but I tested and _ doesnt seem to be used in ksh93. no idea about ksh88 — guest
– guest, Commented Jun 11 at 18:34

KamilCuk · Accepted Answer · 2025-06-11 16:22:43Z

7

Just:

eval "unset _rc; return $_rc"

answered Jun 11 at 16:22

KamilCuk

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9 Comments

Charles Duffy Jun 11 at 17:20

as an aside: If we didn't need ksh compatibility and could restrict this to bash 5.0 and later, I'd be suggesting ${_rc@Q} just to foreclose any possibility that _rc could get a non-integer / attacker-controlled value. Shouldn't ever be relevant if the flow control doesn't allow this to be run other than behind _rc=$?, but extra robustness is extra robustness.

jhnc Jun 11 at 17:30

@CharlesDuffy oops, yeah. I recently came from stackoverflow.com/questions/79656811/… and forgot this isn't escaped

guest Jun 11 at 18:30

this is very clever. why does the variable stick around untill after eval'd though? Also It still complains about returning when the script is not sourced. I will try to use this as the base for the logic I used before

jhnc Jun 11 at 18:59

@guest you could try: eval "unset _rc; (exit $_rc)". I think the answer is meant to be used in the if/return leg of the code in your question

jhnc Jun 11 at 19:00

@guest the variable doesn't stick around. Its value is interpolated into the string. Then eval just sees the constant.

|

jeb · Accepted Answer · 2025-06-12 07:54:21Z

4

You could use this, to inject the _rc value before unsetting it

_rc=42
{
    unset _rc
    return "$(cat)"
} <<< "${_rc}"

Or as oneliner

{ unset _rc; return "$(cat)"; } <<< "${_rc}"

edited Jun 12 at 7:54

answered Jun 11 at 16:01

jeb

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5 Comments

Charles Duffy Jun 11 at 16:08

Better to use echo "$value" -- don't want IFS=1 to change 10 to 0.

psmears Jun 11 at 16:09

@CharlesDuffy: Or maybe return $(cat)?

Charles Duffy Jun 11 at 16:12

Reasonable -- { unset _rc; return "$(cat)"; } <<<"$_rc" should work fine -- though the quotes are still called for around the substitution to prevent word-splitting. Main argument that comes to mind against that is that cat is an external executable to run, so read+echo is going to be faster.

jeb Jun 11 at 16:18

@CharlesDuffy Yes, you are right about IFS. IMHO your $(cat) variant looks best

guest Jun 11 at 19:39

This was a good answer and I +1'd. The eval method worked better for me when _rc was entirely controlled by me and I marked it as a solution but this answer has an advantage over the eval one if you need information form a variable that a user has input to

guest · Accepted Answer · 2025-06-11 19:35:30Z

2

I'm adding this as an extension of @KamilCuk's answer for people like me who needed a little extra info to fill in the gaps. This is the complete solution that works when sourced in bash/ksh and when executed:

#!/bin/bash
# inputs for testing
_rc=2 # for testing

(
  [[ -n $KSH_VERSION && "$(cd -- "$(dirname -- "$0")" && pwd -P)/$(basename -- "$0")" != "$(cd -- "$(dirname -- "${.sh.file}")" && pwd -P)/$(basename -- "${.sh.file}")" ]] || 
  [[ -n $BASH_VERSION ]] && (return 0 2>/dev/null)
) && _was_sourced=1 || _was_sourced=0

#solution
if [ "${_was_sourced}" == 1 ]; then
    unset _was_sourced
    eval "unset _rc; (return $_rc)"
else 
    unset _was_sourced
    eval "unset _rc; (exit $_rc)"
fi

answered Jun 11 at 19:35

guest

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4 Comments

Charles Duffy Jun 11 at 20:02

Aside: When aiming for portability, better to use = than == in [. See pubs.opengroup.org/onlinepubs/9799919799: behavior of [ s1 = s2 ] is standardized; [ s1 == s2 ] is not.

Charles Duffy Jun 11 at 20:07

BTW, you can use (exit $_rc) even when was_sourced is true, because the exit only applies to the subshell that the parens create.

Barmar Jun 11 at 20:34

Why do you need to unset anything in the non-sourced case?

guest Jun 12 at 10:11

@Barmar Its a good point, I dont. I could have just exited with the desired run code in the else statement there

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