Why are the bash -n and -z test operators not inverses for $@

Question

function wtf() {
  echo "\$*='$*'"
  echo "\$@='$@'"
  echo "\$@='"$@"'"
  echo "\$@='""$@""'"
  if [ -n "$*" ]; then echo " [ -n \$* ]"; else echo "![ -n \$* ]"; fi
  if [ -z "$*" ]; then echo " [ -z \$* ]"; else echo "![ -z \$* ]"; fi
  if [ -n "$@" ]; then echo " [ -n \$@ ]"; else echo "![ -n \$@ ]"; fi
  if [ -z "$@" ]; then echo " [ -z \$@ ]"; else echo "![ -z \$@ ]"; fi
}

wtf

produces

$*=''
$@=''
$@=''
$@=''
![ -n $* ]
 [ -z $* ]
 [ -n $@ ]
 [ -z $@ ]

though it seems to me that [-n $@] should be false because 7.3 Other Comparison Operators indicates that [ -n "$X" ] should be the inverse of [ -z "$X" ] for all $X.

-z

string is null, that is, has zero length

String=''   # Zero-length ("null") string variable.

if [ -z "$String" ]
then
  echo "\$String is null."
else
  echo "\$String is NOT null."
fi     # $String is null.

-n

string is not null.

The -n test requires that the string be quoted within the test brackets. Using an unquoted string with ! -z, or even just the unquoted string alone within test brackets (see Example 7-6) normally works, however, this is an unsafe practice. Always quote a tested string. [1]

I know $@ is special but I did not know it was special enough to violate boolean negation. What is happening here?

---

$ bash -version | head -1
GNU bash, version 4.2.42(2)-release (i386-apple-darwin12.2.0)

---

The actual numeric exit codes are all 1 or 0 as per

$ [ -n "$@" ]; echo "$?"
0

Answer 1

When $@ is empty, "$@" doesn't expand to an empty string; it is removed altogether. So your test is not

[ -n "" ]

but rather

[ -n ]

Now -n isn't an operator, but just a non-empty string, which always tests as true.

Answer 2

"$@" doesn't do what you expect. It's not a different form of "$*", it expands to the quoted list of arguments passed to the current script.

If there are no arguments, it expands to nothing. If there are two arguments a and b c, then it expands to "a" "b c" (i.e. it preserves whitespace in arguments) while "$*" expands to "a b c" and $* would expand to a b c (three words).