为什么bash -n和-z测试运算符不能反转$@

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

产生

06001

虽然在我看来[-n $@]应该是假的,因为7.3 Other Comparison Operators表示[-n“$X”]应该是所有$X的[-z“$X”]的倒数.

-z

string is null,that is,has zero length

06002

-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]

我知道$@是特殊的,但我不知道它是否足以违反布尔否定.这里发生了什么？

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

实际的数字退出代码均为1或0

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