Why is $#ARGV 0?

Question

I've been searching through Perl resources, and I could not see where I am going wrong. I am pretty sure I'm missing something obvious, because when I assign

my $gArgc = $#ARGV;

but call my program perl pkTkPtBdTkNo.pl test.txt

$#ARGV is equal to 0, and I can't figure out why.

#! /usr/bin/perl -w

use strict;

my $gArgc = $#ARGV;
my $input_line;
my $bad_input;

print($gArgc);
die ("Usage pkTkPtBdTkNo.pl input-line")
    if(0 == $gArgc);

$input_line = $ARGV[0];
$bad_input = ($input_line =~ /\"\d+\D+\d*\",/);
print($bad_input); 

Answer 1

From man perlintro:

The special variable $#array tells you the index of the last element of an array: ... You might be tempted to use $#array + 1 to tell you how many items there are in an array. Don't bother. As it happens, using @array where Perl expects to find a scalar value ("in scalar context") will give you the number of elements in the array:

So, if you pass 0 arguments, $#ARGV will be -1, since there are no elements in the array. If you pass 1 argument (as in your example), $#ARGV will be 0.

Answer 2

This should be always true, $#ARGV+1 == @ARGV as $#ARGV is last index of @ARGV array.