bash - $# 和 ${#@} 的区别

Question

我在看下面的代码，发现 $# 和 ${#@} 打印出相同的值。谁能告诉我这两者有什么区别？

# length.sh

E_NO_ARGS=65

if [ $# -eq 0 ]  # Must have command-line args to demo script.
then
  echo "Please invoke this script with one or more command-line arguments."
  exit $E_NO_ARGS
fi  

var01=abcdEFGH28ij
echo "var01 = ${var01}"
echo "Length of var01 = ${#var01}"
# Now, let's try embedding a space.
var02="abcd EFGH28ij"
echo "var02 = ${var02}"
echo "Length of var02 = ${#var02}"

echo "Number of command-line arguments passed to script = ${#@}"
echo "Number of command-line arguments passed to script = $#"

exit 0

Answer 1

每个手册页 ( 3.5.3 Shell Parameter Expansion ):

${#parameter}

Parameter length. The length in characters of the value of parameter is substituted. If parameter is * or @, the value substituted is the number of positional parameters. If parameter is an array name subscripted by * or @, the value substituted is the number of elements in the array.

所以 ${#@} 只是一个特例，因为 $# 是表示位置参数数量的常用方式。