What does -n mean in Bash?

Question

This is probably a really dumb question, but... if [ ! -n "$1" ], means, if not more than one argument do... so I get how it works, but what is -n is it the abbreviation of number?

I've been reading through The Advanced Bash programming guide and they just start using it. I've tried to find it and come up with it must be a "built-in" default parameter. Is there a command to show default parameters in Linux?

Answer 1

The -n argument to test (aka [) means "is not empty". The example you posted means "if $1 is not not empty. It's a roundabout way of saying [ -z "$1" ]; ($1 is empty).

You can learn more with help test.

$1 and others ($2, $3..) are positional parameters. They're what was passed as arguments to the script or function you're in. For example, running a script named foo as ./foo bar baz would result in $1 == bar, $2 == baz

Answer 2

-n is one of the string operators for evaluating the expressions in Bash. It tests the string next to it and evaluates it as "True" if string is non empty.

Positional parameters are a series of special variables ($0, $1 through $9) that contain the contents of the command line argument to the program. $0 contains the name of the program and other contains arguments that we pass.

Here, [ ! -n "$1" ] evaluates to "True" if the second positional parameter ($1) passed to the program is empty or (in other words) if no arguments are passed to the program other than $0.