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Perl Display And Pass Command Line Arguments With @argv

How do I read or display command-line arguments with Perl?

Perl command line arguments stored in the special array called @ARGV. The array @ARGV contains the command-line arguments intended for the script. $#ARGV is generally the number of arguments minus one, because $ARGV[0] is the first argument, not the program's command name itself. Please note that $ARGV contains the name of the current file when reading from <>.

Perl @ARGV examples

Use $ARGV[n] to display argument.

Use $#ARGV to get total number of passed argument to a perl script.

For example, if your scriptname is foo.pl:

./foo.pl one two three

You can print one, two, three command line arguments with print command:

print "$ARGV[0]\n";
print "$ARGV[1]\n";
print "$ARGV[2]\n";

Or just use a loop to display all command line args:

# get total arg passed to this script
my $total = $#ARGV + 1;
my $counter = 1;
# get script name
my $scriptname = $0;
print "Total args passed to $scriptname : $total\n";
# Use loop to print all args stored in an array called @ARGV
foreach my $a(@ARGV) {
	print "Arg # $counter : $a\n";

Sample outputs:

./script.pl -c tom -m jerry
Total args passed to /script.pl : 4
Arg # 1 : -c
Arg # 2 : tom
Arg # 3 : -m
Arg # 4 : jerry

More examples

#!/usr/bin/perl -w
if ($#ARGV != 2 ) {
	print "usage: mycal number1 op number2\neg: mycal 5 + 3 OR mycal 5 - 2\n";
if ( $op eq "+" ) {
	$ans = $n1 + $n2;
elsif ( $op eq "-"){
	$ans = $n1 - $n2;
elsif ( $op eq "/"){
	$ans = $n1 / $n2;
elsif ( $op eq "*"){
	$ans = $n1 * $n2;
else {
	print "Error: op must be +, -, *, / only\n";
print "$ans\n";

Save and run script as follows:
$ chmod +x mycal.pl
$ ./mycal.pl
$ ./mycal.pl 5 + 3
#### Note: * need to be escaped under UNIX shell ###
$ ./mycal.pl 5 \* 3

Sample outputs:

Fig.01: mycal.pl in action

Fig.01: mycal.pl in action

{ 15 comments… add one }

  • kunal November 23, 2007, 1:58 pm

    i think u wrongly misspelled the spelling of line to ling in “Perl display and pass command ling arguments with @argv”

  • jaspreet September 6, 2008, 6:51 am

    Helpful info! Thanks!!

  • Chazz December 10, 2008, 7:22 pm

    Here’s a simple way to display your command args:

     foreach (@ARGV) { print "$_\n" }

    or you can use:

      map { print "$_\n" } @ARGV;

    Also, if you want to have command line options such as (-a foo), you can use the getopts perl module. Here’s an example:

    use Getopt::Std;
      getopts ("ab:");
      print "a exists\n" if $opt_a;
      print "b = $opt_b\n" if $opt_b;

    The getopts(…) line specifies the options you want. If a letter ends with “:”, then it can provide a string. otherwise it will be a flag option. If the option exists at command line by user, you will have a variable automatically generated prefixed with $opt_. Here’s an example of the command line for the previous code:

    example.pl -a -b foo

    Would display:

    a exists
    b = foo

    The options must be single letter. For longer, more complex command options, see the Getopt::Long perl module (google is your friend).

    Hope this helps.

  • Brian February 16, 2009, 5:58 pm

    Thanks for proving the information and example for passing command line arguments with $argv. The following line of the calculator script does not work on my system, and I suspect it is related to the “*” metacharacter.

    elsif ( $op eq “*”){
    $ans = $n1 * $n2;

    Being new to Perl scripting, I have to ask if the syntax is correct for this character match? When I execute the script with a multiplication operator, the script returns an error; however, if I change the code to match “x” instead of “*” it runs perfectly.


  • av May 14, 2009, 11:23 am

    Is there a limit to the num,ber of arguments a perl script can take?

  • rt August 12, 2009, 1:07 pm

    The first suggestion

    print “$ARGV[$0]\n”;
    print “$ARGV[$1]\n”;
    print “$ARGV[$2]\n”;

    throws an error (undeclared variable) – it uses a zero (0) for each line.

    The followring works –

    print “$ARGV[0]\n”;
    print “$ARGV[1]\n”;
    print “$ARGV[2]\n”;

  • mac November 3, 2009, 1:19 am

    Hi Brian,

    Your shell ( dos ) is substituting your * argument. you should either escape it with \* when calling the program or use something else, as you did. quotes would work too:
    calc.pl 10 ‘*’ 15


  • Yogesh June 20, 2010, 4:39 am

    Hi, I have a query

    Say for example I have the following in an excel sheet:

    A 1 2 3
    B 4 5 6
    C 7 8 9

    I want to get the data attached to A and say pick up 2 from there and pass it to another perl script irrespective of the cell number in which it is.

    Can u please advise me on that.

  • Virag February 11, 2011, 10:41 pm

    I have a problem with this
    when i do
    path.pl bus[0] bus[1]

    It is unable to take the arguements with the [] braces and the code terminates

  • Vishal May 11, 2011, 4:14 am

    What is the functionlity of ‘$@’ command line argument in Perl?

  • dileep October 20, 2011, 2:44 pm

    how to pass linux commands in perl script?

  • M5J5 January 6, 2012, 7:34 pm

    A couple of corrections if I may:

    print “$ARGV[$0]\n”;
    print “$ARGV[$1]\n”;
    print “$ARGV[$2]\n”;

    should be read

    print “$ARGV[0]\n”;
    print “$ARGV[1]\n”;
    print “$ARGV[2]\n”;

    Command Lin*E* not Ling

    THANKS for this post!!!

  • David April 9, 2012, 5:29 pm

    $#ARGV actually gives you the last index in the @ARGV array, not the total number of arguments passed. scalar(@ARGV) will give you the total number of arguments or just assign to a scalar.
    see http://perldoc.perl.org/perldata.html#Scalar-values for details

    for e.g.

    [root@here]# cat argv.pl
    my $argc = @ARGV;
    my $lastIndex = $#ARGV;
    print "argc=$argc lastIndex=$lastIndex\n";
    [root@here]# ./argv.pl
    argc=0 lastIndex=-1
    [root@here]# ./argv.pl arg1
    argc=1 lastIndex=0
    [root@here]# ./argv.pl arg1 arg2
    argc=2 lastIndex=1
  • Patrick October 26, 2012, 6:55 pm

    Yes, David is correct. $#ARGV is absolutely the WRONG way to get the number of arguments (that’s Bash Script Syntax; not Perl syntax). This post is misleading and should be fixed.

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