Bash, confusing results for different file tests (

Question

I am confused in bash by this expression:

$ var="" # empty var
$ test -f $var; echo $? # test if such file exists
0 # and this file exists, amazing!
$ test -f ""; echo $? # let's try doing it without var
1 # and all ok

I can't understand such bash behaviour, maybe anybody can explain?

Answer 1

When var is empty, $var will behave differently when if quoted or not.

test -f $var          # <=> test -f       ==>   $? is 0
test -f "$var"        # <=> test -f ""    ==>   $? is 1

So this example tells us: we should quote the $var.

Answer 2

It's because the empty expansion of $var is removed before test sees it. You are actually running test -f and thus there's only one arg to test, namely -f. According to POSIX, a single arg like -f is true because it is not empty.

From POSIX test(1) specification:

1 argument:
Exit true (0) if `$1` is not null; otherwise, exit false.

There's never a test for a file with an empty file name. Now with an explicit test -f "" there are two args and -f is recognized as the operator for "test existence of path argument".