Difference between `if [ ! -z "$var" ] then` and `if [ "$var"] then" in bash

Question

Is there any difference between

if [ ! -z "$var" ] then
    # do smth
fi

and

if [ "$var" ] then
    # do smth
fi

They both seem to check if variable is set

-z string True if the string is null (an empty string) -- https://stackoverflow.com/questions/18096670/what-does-z-mean-in-bash — Book Of Zeus, Sep 23 '21 at 00:11
@BookOfZeus That means that `! -z` is true if the string is *not* empty. Which is what `[ "$var" ]` also tests. — Barmar, Sep 23 '21 at 00:13

score 3 · Accepted Answer · answered Sep 23 '21 at 00:23

Yes, they're equivalent, but there are a couple of notes that apply to both of them:

You need either a semicolon or a line break between ] and the then keyword, or it'll misparse them weirdly.
They're not testing whether the variable is set, they're testing whether it's set to something other than the empty string (see this question for ways to check whether a variable is truly unset).

However, I actually prefer a third also-equivalent option:

if [ -n "$var" ]; then

I consider this semantically clearer, because the -n operator specifically checks for something being non-empty.

Ah, thanks for detailed answer! – zovube Sep 23 '21 at 01:18 — zovube, Sep 23 '21 at 01:18

score 1 · Answer 2 · answered Sep 23 '21 at 00:16

1

There's no difference between

[ ! -z "$var" ]
[ -n "$var" ]
[ "$var" ]

All of them are true if $var is not empty.

answered Sep 23 '21 at 00:16

Barmar

2 Answers2