CODEWITHSUNDEEP
bash
Biganon
Biganon

Reputation: 189

Why does 'test -n' return 'true' in bash?

I was wondering how comes

test -n

return 'true', for example :

if test -n; then echo "yes" ; else echo "no" ; fi

prints "yes", even though test was given, theoretically, an empty-length string as an argument along with the option -n, which checks whether the string length is 0 (returns false) or something else (returns true).

Thank you

Upvotes: 4

Views: 5586

Answers (3)

fossildoc
fossildoc

Reputation: 45

Let's be more specific. Consider this script:

if [ -n $myvar ]
then
     echo true
else
     echo false

Notice that myvar is not defined. The script prints "true", which is not expected. Now change $myvar to "$myvar". The script prints "false", which makes sense. Notice also that

test $myvar

also prints false, as expected. This difference is not explained anywhere in the bash docs or numerous SO posts on this observation. The quotes are supposed to make bash treat the variable contents as a single string, but that is not applicable here. Any help, please.

Upvotes: 0

Paŭlo Ebermann
Paŭlo Ebermann

Reputation: 74800

From the documentation:

The test and [ builtins evaluate conditional expressions using a set of rules based on the number of arguments.

0 arguments: The expression is false.

1 argument: The expression is true if and only if the argument is not null.

In your case you simply have one non-null argument (-n).

Upvotes: 5

Jonathan Leffler
Jonathan Leffler

Reputation: 754560

It returns true for the same reason test x returns true - the string -n is non-empty. It is not exercising the -n option because -n requires a second argument and you've not provided one.

test -n  ""  || echo false
x=""
test -n  $x  && echo true
test -n "$x" || echo false

Each echo command is executed; note, in particular, the middle one!

Upvotes: 3

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