What does ${##} mean in bash?

Question

I know $# is the number of positional parameters in bash. But how does bash interpret ${##}? Here, is a sample output from my system.

$ echo $#
0
$ echo ${#}
0
$ echo ${##}
1
$ echo $##
0#

Answer 1

$# is the number of positional parameters, and ${##} is the length of $#'s value in characters; in other words it's the base 10 logarithm plus one, rounded down, of the number of positional parameters. $## doesn't work because it doesn't comply with the parameter expansion syntax.

Observe:

$ bash -c 'echo "$# ${##}"' _ {1..9}
9 1
$ bash -c 'echo "$# ${##}"' _ {1..10}
10 2
$ bash -c 'echo "$# ${##}"' _ {1..100}
100 3

See Bash Reference Manual § 3.5.3 Shell Parameter Expansion for further information.