It seems I am always learning something new about Perl. Today while debugging some graph navigation code I discovered something that surprised me a bit. I was naively using eq to see if we had visited a graph node before, but for some reason my code was insisting that we had already visited a node that I knew we hadn't visited before. It turned out that the reason for this confusing behavior was that qr{someregex} eq '(?-xism:someregex)' was returning true. For example, the following code:

use strict;
use warnings;

my $s='(?-xism:a)';
my $re=qr{a};

# regex and string are clearly different ref types
print "-------Type------------\n";
print "ref $re (regex) is 'Regexp': "
  , (ref($re) eq 'Regexp' ?'true':'false'), "\n";
print "ref $s (string) is '': "
  , (ref($s) eq '' ?'true':'false'), "\n";
print "ref $s (string) is not 'Regexp': "
  , (ref($s) ne 'Regexp' ?'true':'false'), "\n";


# so why are they equal?
print "-------Equality--------\n";
print "comparing literals: qr{a} eq '(?-xism:a)': "
  , (qr{a} eq '(?-xism:a)'?'true':'false'), "\n";
print "comparing variables: regex eq $s (string): "
  , ($s eq $re?'true':'false'), "\n";
[download]

prints

-------Type------------
ref (?-xism:a) (regex) is 'Regexp': true
ref (?-xism:a) (string) is '': true
ref (?-xism:a) (string) is not 'Regexp': true
-------Equality--------
comparing literals: qr{a} eq '(?-xism:a)': true
comparing variables: regex eq (?-xism:a) (string): true
[download]

I'm used to eq comparing numbers to strings - Perl considers them both to be scalars, but here Perl was considering two things that were clearly different types as "equal" - one a reference to a regex and the other a string. So my questions are:

• For what other pairings of data types does eq ignore type?
• Is there a way to override this so that things belonging to different data types are always not equal? Or is the rather verbose (ref($x) eq ref($y)) and ($x eq $y) the only way to do this?
• I know there is a way to overload operators but I was under the impression that one had to "use overload" to empower it. Is that wrong? If so, how can I detect overloaded operators?
• Is the ability to overload operators in any way connected to this behavior of eq? If not, how should I understand it?

Best, beth