unable to give value to a scalar from command line in perl - perl

Question

I am trying the following perl code:
use strict;
use warnings;
use Scalar::Util ("looks_like_number");
my $color;
if (undef $color)
{
my $color = $ARGV[0];
}
print "$0\n";
print "$ARGV[0]\n";
my #colors = ("blue", "yellow", "brown", "white");
print "Please select a num:\n";
foreach my $i (0..$#colors)
{
my $j = $i+1;
print " $j $colors[$i]\n";
}
my $num = <STDIN>;
chomp($num);
if (looks_like_number($num) and defined $colors[$num-1])
{
$color = $colors[$num-1];
}
else
{
print "Bad Selection\n";
}
print "selected color is $color\n";
I want to select any number for the corresponding color choice or I should be able to provide a value of color by $color variable through command line, I am trying to run it in windows cmd using ' perl [C:/scriptname.pl] [color] ' but its not taking the argument but when I am printing ARGV[0] it is showing the argument being passed correctly. so what is the issue with my 'if (undef ARGV[0])' statement that its not getting executed.

Answer 1

You are declaring the variable $color twice:
my $color;
if (undef $color)
{
my $color = $ARGV[0];
}
The second my $color will create a second binding of the name $color and the value you assign to it and that binding will not be visibile outside of the scope, the enclosing curly brackets.
The expression
if (undef $color)
does not do what you intend it to do. undef will always set a value to undef. You want to use defined instead.
After applying those two changes, the code could look like:
my $color;
if (! defined $color)
{
$color = $ARGV[0];
}

Answer 2

my $color;
if (undef $color)
{
my $color = $ARGV[0];
}
This looks rather strange. In line 1 you declare a scalar variable called $color. In line 2, you call undef() on that variable which replaces the contents of $color with undef() (that's unnecessary as a newly-declared Perl scalar will always contain undef()). This expression will return undef, which is false, so the code in your if block is never executed.
In line 4, you declare a new variable, also called $color and set that to $ARGV[0]. There are two problems with this. Firstly, that line will never be executed for the reasons explained in the previous paragraph. And, secondly, your new $color variable will cease to exist once you leave the block, so you would never see the effect of this change.
I think the main problem here is that you have confused undef() with defined(). undef() will always give a variable an undefined values, but I suspect you wanted defined() which tells you whether or not a variable contains a defined value.
What you probably wanted was:
my $color;
if (!defined $color) {
$color = $ARGV[0];
}
Which can be written as:
my $color //= $ARGV[0];
But it's still slightly confusing as it's not clear why you think $color could ever be defined immediately after being declared.