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 ar

Answer 1

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

Answer 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.

Answer 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

Answer 4

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