bash if -a vs -e option
BashBash Problem Overview
Both about -a
and -e
options in Bash documentation is said:
-a file
True if file exists.
-e file
True if file exists.
Trying to get what the difference is I ran the following script:
resin_dir=/Test/Resin_wheleph/Results
if [ -e ${resin_dir} ] ; then
echo "-e ";
fi
if [ ! -e ${resin_dir} ] ; then
echo "! -e";
fi
if [ -a ${resin_dir} ] ; then
echo "-a";
fi
if [ ! -a ${resin_dir} ] ; then
echo "! -a";
fi
/Test/Resin_wheleph/Results
exists and is a directory. And this is what I get:
-e
-a
! -a
which seems to be a little strange (notice -a
and ! -a
). But when I use double brackets (e. g. if [[ -e ${resin_dir} ]]
) in the similar script it gives reasonable output:
-e
-a
So:
- What is a difference between
-a
and-e
options? - Why
-a
produces a strange result when used inside single brackets?
Bash Solutions
Solution 1 - Bash
I researched, and this is quite hairy:
-a
is deprecated, thus isn't listed in the manpage for /usr/bin/test
anymore, but still in the one for bash. Use -e
. For single '[', the bash builtin behaves the same as the test
bash builtin, which behaves the same as /usr/bin/[
and /usr/bin/test
(the one is a symlink to the other). Note the effect of -a
depends on its position: If it's at the start, it means file exists
. If it's in the middle of two expressions, it means logical and
.
[ ! -a /path ] && echo exists
doesn't work, as the bash manual points out that -a
is considered a binary operator there, and so the above isn't parsed as a negate -a ..
but as a if '!' and '/path' is true
(non-empty). Thus, your script always outputs "-a"
(which actually tests for files), and "! -a"
which actually is a binary and
here.
For [[
, -a
isn't used as a binary and
anymore (&&
is used there), so its unique purpose is to check for a file there (although being deprecated). So, negation actually does what you expect.
Solution 2 - Bash
The '-a
' option to the test operator has one meaning as a unary operator and another as a binary operator. As a binary operator, it is the 'and' connective (and '-o
' is the 'or' connective). As a unary operator, it apparently tests for a file's existence.
The autoconf
system advises you to avoid using '-a
' because it causes confusion; now I see why. Indeed, in portable shell programming, it is best to combine the conditions with '&&
' or '||
'.
I think @litb is on the right track. When you have '! -a ${resin_dir}
', Bash may be interpreting it as "is the string '!' non-empty and is the string in '${resin_dir}' non-empty, to which the answer is yes. The Korn shell has a different view on this, and the Bourne shell yet another view - so stay away from '-a
'.
On Solaris 10:
$ bash -c 'x=""; if [ ! -a "$x" ] ; then echo OK ; else echo Bad; fi'
Bad
$ ksh -c 'x=""; if [ ! -a "$x" ] ; then echo OK ; else echo Bad; fi'
OK
$ sh -c 'x=""; if [ ! -a "$x" ] ; then echo OK ; else echo Bad; fi'
sh: test: argument expected
$
Solution 3 - Bash
The double bracket [[ exp ]] is a bash builtin. In bash -a and -e are the same, probably for some backwards compatibility.
The single bracket [ exp ] is an alias for the external command "test". In "test", -a is a logical AND. Although [ nothing AND $STRING ] looks like it should be false, test has some syntax quirks, which is why I recommend using the bash builtin [[ exp ]], which tends to be more sane.
Note: bash really does call /bin/[ when you use "[".
$ [ $UNASIGNED_VAR == "bar" ]
bash: [: ==: unary operator expected
the error shows bash called [. An strace also shows "execve("/usr/bin/[", ..."