Because if $x is an object, then your regex won't work. (Nevermind that it won't work for reals or negative numbers.)

A stringification method (which your regex assumes) isn't reliable, because I could write a stringification method to output the number in hexidecimal, Base64, or as a roman numeral (see below) but as far as Perl was concerned, it could still operate just like a number):

package RomanNumber;

use Roman 'roman';

sub new {
  my $class = shift;
  my $self  = { value => shift };
  bless $self, $class;
}

sub as_num {
  my $self = shift;
  return $self->{value};
}

sub as_string {
  my $self = shift;
  return roman($self->as_num);
}

sub compare {
  my ($a,$b) = @_;
  return ($a->as_num <=> $b->as_num);
}


use overload
  '0+'  => \&as_num,
  '""'  => \&as_string,
  '<=>' => \&compare;

[download]

If I add a few more numeric operators (it's an incomplete example), Perl can treat it just like a number, except that it outputs a roman numeral in string context.

There are cases where I would like to require a value to be number, but I would like to give the user of a module the ability to use a 'number-like' object rather than a number.

In reply to Re: I don't understand. by rrwo
in thread Detecting if a scalar has a number or string by rrwo

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