Script question

[Nobody's Hero]

What does '$#' represent ?

I know '$?' is return code. Not sure about '$#'

[A. Clay Stephenson · Accepted solution]

It's the number of arguments. Inside a function, it's the number of arguments passed to the function otherwise it's the number of argument passed on the command line. Note: The "shift" command, reduces ${#} so after doing a shift 2, for example, ${#} is now 2 less than the number actually passed.

[Mel Burslan]

number of arguments after the command on the command line is represented by $#

command parm1 parm2 parm3

if in the command, you have a line like echo $#, you will see 3 as output to the command line above

Hope this helps

[Sandman!]

It's the number of arguments passed to a script excluding any command line options (which need to be parsed with the getopts command). Within the script the command line arguments equate to the positional parameters $1, $2, $3 all the way upto $9; with $0 being the name of the script itself.

So if your script is named "script.ksh" and it takes arguments and options like...

# script.ksh argument1 argument2 -o argument3
| | | |
$0 $1 $2 $3

hope this helps!

[Basheer_2]

$# or ${#} is the count of args

eg:
cmd arg1 arg2 arg3

${#} = 3
${1} = arg1
${2} = arg2
${3} = arg3
${*} = "arg1 arg2 arg3"
${@} = "arg1" "arg2" "arg3"