why in bash shell # sign instead of $ code example

Example: bash special dollar sign shell varaibles

$1, $2, $3, ... are the positional parameters
"$@" is an array-like construct of all positional parameters, {$1, $2, $3 ...}.
"$*" is the IFS expansion of all positional parameters, $1 $2 $3 ....
$# is the number of positional parameters.
$- current options set for the shell.
$$ pid of the current shell (not subshell).
$_ most recent parameter (or the abs path of the command to start the current shell immediately after startup).
$IFS is the (input) field separator.
$? is the most recent foreground pipeline exit status.
$! is the PID of the most recent background command.
$0 is the name of the shell or shell script.