What is the use of $# in Bash

user_5404687 0
Viewed 20204

I am very new to Bash scripting, can someone explain to me how the $# and $? work in the following code? 

#!/bin/bash

ARGS=3         # Script requires 3 arguments.
E_BADARGS=85   # Wrong number of arguments passed to script.

if [ $# -ne "$ARGS" ]
then
  echo "Usage: `basename $0` old-pattern new-pattern filename"
  exit $E_BADARGS
fi

old_pattern=$1
new_pattern=$2

if [ -f "$3" ]
then
    file_name=$3
else
    echo "File \"$3\" does not exist."
    exit $E_BADARGS
fi

exit $?
linux bash
4 Answers

In bash exist special variables... and i write you some of then.

$#- this is an special variable that content inside the number of command line (you can just count how many parameters were entered) you passed to the script. tis variable also represent the last command line but its better do this ${!#}

$?- this one is very special cause its represents is your script is fine this variable holds the exit status of the previosly command... its a littler confusing but it work perfectly... when you end you script you can positional this variable at the end and if she return 0 value you scrip is perfect is true, if she return 1 or others you must check out your lines.

user_151359690
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