Bash null binary operators

Question

A recent test I took had a question on the output of the following bash command:

var=; [ -n $var ]; echo $?; [ -z $var ]; echo $?

The results are 0 and 0, indicating the return codes for both unary operators had no errors. This means $var resolves to both null (empty) and 'non-null' (not empty), correct?

How is this possible?

Those are unary operators.

Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams

2013-02-16 04:12:36 +00:00
Commented Feb 16, 2013 at 4:12 — Ignacio Vazquez-Abrams
– Ignacio Vazquez-Abrams, Commented Feb 16, 2013 at 4:12
info coreutils test invocation 16.3.4 String tests

tijagi
– tijagi

2013-02-16 07:39:44 +00:00
Commented Feb 16, 2013 at 7:39 — tijagi
– tijagi, Commented Feb 16, 2013 at 7:39

Ignacio Vazquez-Abrams · Accepted Answer · 2013-02-16 04:14:32Z

3

No, it means that [ is unfixably broken. In both cases $var evaluates to nothing, and the commands simply execute [ -n ] and [ -z ] respectively, both of which result in true. If you want to test the value in the variable itself then you must quote it to have it handled properly.

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

answered Feb 16, 2013 at 4:14

Ignacio Vazquez-Abrams

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1 Comment

William Pursell Over a year ago

It is not unfixably broken; it just has non-intuitive behavior. [ -n ] correctly reports that the string -n is not null, and [ -z ] correctly reports that the string -z is not null.

Hai Vu · Accepted Answer · 2013-02-16 04:18:20Z

2

You will need to surround $var:

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

Remember that the closing square bracket is just syntactic sugar: you don't need it. That means your line:

$ [ -n $var ]; echo $?

will expand to (since $var is empty):

$ [ -n  ]; echo $?

The above asks: "is the string ']' non-empty?" And the answer is yes.

answered Feb 16, 2013 at 4:18

Hai Vu

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1 Comment

chepner Over a year ago

While [ is not syntax and in fact a synonym for the test command, you do need to use the closing ] with [.

0xC0000022L · Accepted Answer · 2013-02-16 04:15:01Z

1

It's surprising indeed. If you were to try the same with the bashism [[ syntax, you'd get 1 and 0 as results. I reckon this is a bug.

var=; [[ -n $var ]]; echo $?; [[ -z $var ]]; echo $?

or, as Ignacio points out and as in fact I have always been doing intuitively, with defensive coding and quoting:

var=; [[ -n "$var" ]]; echo $?; [[ -z "$var" ]]; echo $?

It's surprising to me that [ behaves this way, because it's a builtin:

$ type [
[ is a shell builtin

Just did a little test and the system command [ behaves in the same broken way as the builtin. So probably it's buggy for compatibility:

var=; /usr/bin/\[ -n $var ]; echo $?; /usr/bin/\[ -z $var ]; echo $?

answered Feb 16, 2013 at 4:15

0xC0000022L

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2 Comments

Ignacio Vazquez-Abrams Over a year ago

[ has a lot of legacy behavior which cannot be modified, due to scripts depending on the broken-by-default behavior. Kind of like the Windows API.

0xC0000022L Over a year ago

@Ignacio Vazquez-Abrams: just noticed. Interesting.

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