If I want to perform some commands given variables aren't set I'm using:

if [[ -z "$a" || -z "$v" ]]
  echo "a or b are not set"

Yet the same syntax doesn't work with -v, I have to use:

if [[ -v a && -v b ]]
  echo "a & b are set"

What is the history behind this? I don't understand why the syntax wouldn't be the same. I've read that -v is a somewhat recent addition to bash (4.2)?

  • What's the question? Are you asking why you need to use the dollar sign?
    – choroba
    Oct 6, 2017 at 15:02
  • a variable can be set to ""
    – Kevin
    Oct 6, 2017 at 21:21

2 Answers 2


Test operators -v and -z are just not the same.

Operator -z tells if a string is empty. So it is true that [[ -z "$a" ]] will give a good approximation of "variable a is unset", but not a perfect one:

  • the expression will yield true if a is set to the empty string rather than unset;

  • the enclosing script will fail if a is unset and the option nounset is enabled.

On the other hand, -v a will be exactly "variable a is set", even in edge cases. It should be clear that passing $a rather than a to -v would not be right, as it would expand that possibly-unset variable before the test operator sees it; so it has to be part of that operator's task to inspect that variable, pointed to by its name, and tell whether it is set.

  • When you say edge case do you mean they handle empty differently? For example a=""? Oct 6, 2017 at 15:07
  • 4
    Note that there are edge cases for [[ -v ]] as well. Like for a variable array type, [[ -v var ]] returns true only if ${var[0]} (in effect same as $var) is set. For associative array, ${var[0]} as well (not ${var[00]} for instance). [[ -v var[@] ]] would return true if the array has any value set (would return false for a variable set as var=()). That's down to the crazy way variable typing is done in ksh/bash. See zsh for a less confusing variable typing. Oct 6, 2017 at 15:19
  • 2
    More succinctly: -z tests a value, while -a tests a name.
    – chepner
    Oct 6, 2017 at 18:12
  • 2
    @chepner, do you mean -v tests a name?
    – tedyyu
    Dec 1, 2020 at 6:51
  • 1
    @tedyyu Oof, yes. -z tests a value, -v tests a name.
    – chepner
    Dec 1, 2020 at 13:51

There's a difference in the meaning of -z and -v:

echo Empty:
x=""  # Same with x=
[[ -z $x ]] && echo z
[[ -v  x ]] && echo v
unset x

echo Unset
[[ -z $x ]] && echo z
[[ -v  x ]] && echo v

By using -z, you can't distinguish a variable that was assigned an empty value from a variable that hasn't been assigned any value.

Also, [[ -z $x ]] is still sensible to set -u, while [[ -v x ]] isn't.

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