An empty string is defined, but evaluates to false.

From perldoc -f defined:

' Note: Many folks tend to overuse "defined", and then are surprised to discover that the number 0 and "" (the zero-length string) are, in fact, defined values.'

That's a choice quote. Put another way, the OP is suffering from some confusion that if (defined $x) is intended to be the same as if ($x). In fact, they serve different purposes.

defined tests whether a variable has any value -- a false value is still a value.

Short demonstration code:

my $x;
demonstrate();

$x = 1;
demonstrate();

$x = 0;
demonstrate();

sub demonstrate {
  if (defined $x) { print "\$x has value $x\n"; }
  else { print "\$x is undefined\n"; }

  if ($x) { print "\$x evaluates as TRUE\n"; }
  else { print "\$x evaluates as FALSE\n"; }

  print "\n";
}
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This produces the following output:

$x is undefined
$x evaluates as FALSE

$x has value 1
$x evaluates as TRUE

$x has value 0
$x evaluates as FALSE
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<–radiant.matrix–>
Ramblings and references
The Code that can be seen is not the true Code
I haven't found a problem yet that can't be solved by a well-placed trebuchet

Apparently, in Lisp, this is referred to as a semi-predicate problem.

The very fact that you are using


    if ($2)
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followed by

    $price = $1 . "." . $2
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shows that you are using $2 as a semi-predicate, i.e. to show validity and to hold value.

Next time, whenever you use if ($var), you should consider if $var needs to be a full-predicate, i.e. strictly a boolean to only show true-or-false.

Here is an example to illustrate the point:

#!/usr/bin/env perl
use warnings;
use strict;

my $str;

$str = 'abc def 12';
$str =~ /(\d+)/;
print "$1\n" if ($1);

$str = 'foo';
$str =~ /(\d+)/;
print "$1\n" if ($1);

if ($str =~ /(\d+)/) {
    print "$1\n";
}
else {
    print "no match\n";
}
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prints:

12
12
no match
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The 2nd "12" is unexpected since the string 'foo' contains no digits.