CODEWITHSUNDEEP
bashshell
Kearney Taaffe
Kearney Taaffe

Reputation: 695

what does [ -n "$VARIABLE" ] || exit 0 mean

Looking at correcting an issue in /etc/init.d/hostapd on Debian. However, I have no clue what this line of code does nor how it works

[ -n "$DAEMON_CONF" ] || exit 0

In searching online for bash tutorials, I've never seen anyone do this

When I run the code, my shell window closes (because $DAEMON_CONF is not set to anything). If I change the code to 

[ -n "not empty" ] || exit 0

my console window does not close.

so, -n evaluates to true, and or'ed with exit 0, is what?

Upvotes: 6

Views: 2362

Answers (4)

Mahesh
Mahesh

Reputation: 1683

It checks if the environment variable is defined, if $DAEMON_CONF is not present the it will exit with 0 code, a better code would be.

[ -n "$DAEMON_CONF" ] || echo "exiting as DAEMON_CONF is not set" && exit 1

Upvotes: 0

agc
agc

Reputation: 8406

[ -n "$DAEMON_CONF" ] || exit 0

It's an unnecessary double negative. This would do the same thing:

[ -z "$DAEMON_CONF" ] && exit 0

Or it could be done without any flag:

[ "$DAEMON_CONF" ] || exit 0

Upvotes: 3

JNevill
JNevill

Reputation: 50200

[ is and alternate name for the command test. You can learn about the parameters/flags whatnot by looking at test's manpage:

man test

You'll see for -n:

-n STRING

          the length of STRING is nonzero

Furthemore || means OR. So if the test command returns False then the stuff after the || will be executed. If test returns true, then it won't be executed.

Written out your command says: "If the variable $DAEMON_CONF lacks a value, then exit with return code 0"

The longhand version would be something like:

if test ! -n "$DAEMON_CONF"; then
    exit 0
fi

Upvotes: 4

Elliott Frisch
Elliott Frisch

Reputation: 201467

If the expression in [] returns false, do the thing after the or || (and exit 0). Otherwise, it will short circuit and the next statement will be evaluated.

Upvotes: 7

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