Re: ARGV behaviour in getopts std

but why $#ARGV !=6 is true?

Because $#ARGV is 5, which you can prove by printing its value:

use warnings;
use strict;
use Data::Dumper;
use Getopt::Std;

print '$#ARGV=',  $#ARGV, "\n";
print Dumper(\@ARGV);

if ($#ARGV == -1 || $#ARGV !=6) {
    print "ARGV is now reduced to <".@ARGV."> members:\n";#prints ARGV
+ as 6
}

__END__

$#ARGV=5
$VAR1 = [
          '-b',
          'opt_b',
          '-p',
          'opt_p',
          '-c',
          'opt_c'
        ];
ARGV is now reduced to <6> members:
[download]

@ARGV

Comment on Re: ARGV behaviour in getopts std Select or Download Code

Replies are listed 'Best First'.
Re^2: ARGV behaviour in getopts std by perl_mystery (Beadle) on Dec 21, 2010 at 23:05 UTC
What is the difference between $#ARGV and @ARGV?I can see @ARGV has members,but what is $#ARGV,what does it contain?why is the number five?is it always "1" less than @ARGV?	[reply]
Re^3: ARGV behaviour in getopts std by toolic (Bishop) on Dec 22, 2010 at 01:10 UTC
You will find the answer if you follow the link that I provided in my original reply and read it. If there is something unclear in the documentation, please specify what you don't understand. Here it is again: @ARGV	[reply]
Re^4: ARGV behaviour in getopts std by perl_mystery (Beadle) on Dec 22, 2010 at 02:33 UTC
Cool thanks,couldnt see the link before	[reply]
Re^3: ARGV behaviour in getopts std by Argel (Prior) on Dec 22, 2010 at 17:57 UTC
'scalar @array' is the number of elements in @array (sometimes also referred to as the length of @array). $#array is the last index of @array. By default the first index in an array is 0, so when using the default settings $#array will be 'scalar @array - 1'. Elda Taluta; Sarks Sark; Ark Arks	[reply]
A reply falls below the community's threshold of quality. You may see it by logging in.