What does the unix $# command do?

Question

I read this page What are the special dollar sign shell variables? but I still don't get what $# does.

I have an example from a lecture slide:

#!/bin/sh
echo Total no. of inputs: $#
echo first: $1
echo second: $2

I assume $# takes in all inputs as the argument, and we should expect two inputs. Is that what it's doing?

Answer 1

as your script says: Its total number of command line arguments you are passing to your script.

if you have a script name as: kl.sh

execute it as

./kl.sh hj jl lk or even bash kl.sh hj jl lk

and in script you are doing

echo $#

It will print 3

where

$1 is hj
$2 is jl
$3 is lk

This tutorial will surely help you

Answer 2

$# is a special built-in variable that holds the number of arguments passed to the script.

Using the suggested code for example:

#!/bin/sh
echo Total no. of *arguments*: $#
echo first: $1
echo second: $2

If this script is saved to a file, say, printArgCnt.sh, and the executable permissions of printArgCnt.sh are set, then we can expect the following results using $#:

>> ./printArgCnt.sh A B C 
Total no. of *arguments*: 3
first: A
second: B

>> ./printArgCnt.sh C B A
Total no. of *arguments*: 3   (<-- still three...argument count is still 3)
first: C
second: B