Using unset vs. setting a variable to empty

Question

I'm currently writing a bash testing framework, where in a test function, both standard bash tests ([[) as well as predefined matchers can be used. Matchers are wrappers to '[[' and besides returning a return code, set some meaningful message saying what was expected.

Example:

string_equals() {
    if [[ ! $1 = $2 ]]; then
            error_message="Expected '$1' to be '$2'."

            return 1
    fi
}

So, when a matcher is used, and it fails, only then an error_message is set.

Now, at some point later, I test whether the tests succeeded. If it succeeded, I print the expectation in green, if it failed in red.

Furthermore, there may be an error_message set, so I test if a message exists, print it, and then unset it (because the following test may not set an error_message):

if [[ $error_message ]]; then
    printf '%s\n' "$error_message"

    unset -v error_message
fi

Now my question is, if it is better to unset the variable, or to just set it to '', like

error_message=''

Which one is better? Does it actually make a difference? Or maybe should I have an additional flag indicating that the message was set?

If you never compare error_message with anything else, I'd say it doesn't matter. However, I think you want [[ $error_message ]], otherwise you are testing that the literal string "error_message" exists. — chepner
– chepner, Commented Sep 4, 2012 at 12:44

cdarke · Accepted Answer · 2018-01-21 19:48:08Z

162

Mostly you don't see a difference, unless you are using set -u:

/home/user1> var=""
/home/user1> echo $var

/home/user1> set -u
/home/user1> echo $var

/home/user1> unset var
/home/user1> echo $var
-bash: var: unbound variable

So really, it depends on how you are going to test the variable.

I will add that my preferred way of testing if it is set is:

[[ -n $var ]]  # True if the length of $var is non-zero

or

[[ -z $var ]]  # True if zero length

edited Jan 21, 2018 at 19:48

answered Sep 4, 2012 at 12:54

cdarke

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7 Comments

chepner Over a year ago

var= is not "not set". It's just an unquoted empty string. In addition to set -u, bash's various forms of parameter expansion can also distinguish between unset and null values: ${foo:bar} expands to "bar" when foo is unset, but "" when foo is null, while ${foo:-bar} expands to "bar" if foo is unset or null.

chepner Over a year ago

[[ -n $var ]] is false if var is set to the empty string.

osirisgothra Over a year ago

if you use 'declare -p' to check for variable 'existence' then var= will show 'declare -- var=""' you must use unset to get rid of it, of course if its a readonly variable then you cant get rid of it of course. Furthermore, if you var=something inside a function, you wont have to worry about getting rid of it if you use "local var=value" be careful though, because "declare var=value" is also local.In the case of declare you have to explicitly set it global using "declare -g var=value", without declare you have to explicitly set it to be local using "local".Global vars are only erased /w unset.

cdarke Over a year ago

@osirisgothra: good point about local variables. Of course generally it is best to declare (or localise) all variables in a function so that it is encapsulated.

Liviu Chircu Over a year ago

@chepner should be ${foo-bar} instead of ${foo:bar}. test for yourself: unset a; echo ">${a:-foo}-${a:foo}-${a-foo}<"

|

Zombo · Accepted Answer · 2014-10-18 16:52:01Z

19

As has been said, using unset is different with arrays as well

$ foo=(4 5 6)

$ foo[2]=

$ echo ${#foo[*]}
3

$ unset foo[2]

$ echo ${#foo[*]}
2

answered Oct 18, 2014 at 16:52

Zombo

1

Comments

Jonathan H · Accepted Answer · 2025-01-15 17:44:23Z

3

Based on the comments above, here is a simple test:

isunset() { [[ "${!1}" != 'x' ]] && [[ "${!1-x}" == 'x' ]] && echo 1; }
isset()   { [ -z "$(isunset "$1")" ] && echo 1; }

Example:

$ unset foo; [[ $(isunset foo) ]] && echo "It's unset" || echo "It's set"
It's unset
$ foo=     ; [[ $(isunset foo) ]] && echo "It's unset" || echo "It's set"
It's set
$ foo=bar  ; [[ $(isunset foo) ]] && echo "It's unset" || echo "It's set"
It's set

Here is a suggested improvement:

isset()   { [[ $(eval echo "\${${1}+1}") ]]; }
isunset() { ! isset "$1"; }

Example:

$ unset foo; isunset foo && echo "It's unset" || echo "It's set"
    It's unset
$ foo=     ; isunset foo && echo "It's unset" || echo "It's set"
    It's set
$ foo=bar  ; isunset foo && echo "It's unset" || echo "It's set"
    It's set

edited Jan 15 at 17:44

answered Jul 19, 2018 at 14:37

Jonathan H

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1 Comment

DKebler Jan 11 at 22:10

isunset() { stmp=$(mktemp -u XXXXXXXXXXXXXX); [[ "${!1-$stmp}" == "$stmp" ]] || return 1; } isset() { if isunset "$1"; then return 1; fi }

Paul-Beyond · Accepted Answer · 2018-04-12 07:52:12Z

2

So, by unset'ting the array index 2, you essentially remove that element in the array and decrement the array size (?).

I made my own test..

foo=(5 6 8)
echo ${#foo[*]}
unset foo
echo ${#foo[*]}

Which results in..

3
0

So just to clarify that unset'ting the entire array will in fact remove it entirely.

answered Apr 12, 2018 at 7:52

Paul-Beyond

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