To what does an unquoted empty variable expand in bash?

Question

To what value does an unquoted, undeclared variable expand to in order to return an exit status of 0?

Here is an example of a particular situation I ran into:

[ -n $var ]; echo $?
0
[ -n "$var" ]; echo $?
1

In both tests, the variable var is not declared. I could have saved me the hassle by testing with -z, where quoted or unquoted apparently doesn't make a difference, but I ran into this particular situation and I started wondering. I looked deeper into all the expansions that bash performs, but couldn't find any explanation for this behavior.

As a general rule I usually quote variables, but hopefully the reason of this behavior helps me better understand quoting.

[ -n ] is true (0) since -n is a non-empty string. Also note that [ -z $var ] would have been true for the same reason. I suspect there would be a duplicate question for this somewhere... — Kusalananda, Oct 21 '20 at 12:03
In your case, you can do without the option -n: [ "$var" ] or [ $var ] — nezabudka, Oct 21 '20 at 12:17
Related: How does bash interpret the equal operator with no surrounding spaces in a conditional? — steeldriver, Oct 21 '20 at 13:17
@nezabudka, no, please don't tell people to use (the unquoted) [ $var ], it breaks the moment $var contains more than one word — ilkkachu, Oct 21 '20 at 13:22
On the other hand, [ -z $var ] works quoted or not as long as $var contains one or zero words. I think there was a post explaining that somewhere, but I can't find it so I'll leave this as an exercise... — ilkkachu, Oct 21 '20 at 13:24
@ikkachu, Will it be okay?: [ -n $var ] or [ -z $var ]. "In your case...". Considering a certain example. — nezabudka, Oct 21 '20 at 13:39

score 5 · Accepted Answer · answered Oct 21 '20 at 13:02

The test, [ and [[ commands all act the same in that they do different things based on how many arguments they are given (excluding the closing ]/]]).

with 1 argument, the test will be successful if the argument is not empty
with 2 arguments, the arguments are treated as a unary operator (such as -n) and an operand.
with 3 arguments, the arguments are treated as an operand, a binary operator (such as =) and another operand
more than 3 arguments, test and [ give a "too many arguments" error, and [[ may give a different error (it has a more complicated parser).

Now, looking at the [ -n $var ] example, and comparing against the [[ construct. Since [[ does not do word splitting, it knows where variable values are. When var="", given [[ -n $var ]], after parameter expansion, bash will see [[ -n "" ]] which is clearly false.

But for test and [, word splitting is in effect. So [ -n $var ] becomes [ -n ] (the empty string disappears). Now, [ sees one argument which is a non-empty string, and is therefore true.

If you quote the variable: [ -n "$var" ], then you'll get the expected false result.

There are similar confusing results when var='*' due to filename expansion. — glenn jackman, Oct 21 '20 at 13:04
Thanks, helpful! Also because you indicate that the test behaves differently depending on the number of arguments. — twan163, Oct 21 '20 at 13:31
Also, refer to the canonical Q&A Security implications of forgetting to quote a variable in bash/POSIX shells for ALL the details. — glenn jackman, Oct 21 '20 at 17:45

To what does an unquoted empty variable expand in bash?

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