6 SC2070

Joachim Ansorg edited this page 2021-11-15 11:27:55 +01:00

Table of Contents

• -n doesn't work with unquoted arguments. Quote or use [[ ]].
• Problematic code:
• Correct code:
• Rationale:
• Exceptions:

-n doesn't work with unquoted arguments. Quote or use [[ ]].

Problematic code:

if [ -n $var ]
then
  echo "var has a value"
else
  echo "var is empty"
fi

Correct code:

In POSIX:

if [ -n "$var" ]
then
  echo "var has a value"
else
  echo "var is empty"
fi

In bash/ksh:

if [[ -n $var ]]
then
  echo "var has a value"
else
  echo "var is empty"
fi

Rationale:

When $var is unquoted, a blank value will cause it to wordsplit and disappear. If $var is empty, these two statements are identical:

[ -n $var ]
[ -n ]

[ string ] is shorthand for testing if a string is empty. This is still true if string happens to be -n. [ -n ] is therefore true, and by extension so is [ -n $var ].

To fix this, either quote the variable, or (if your shell supports it) use [[ -n $var ]] which generally has fewer caveats than [.

Exceptions:

None

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