Empty pattern and /o

While looking at an old bug report I was, at first, confused by the  
following code:


#!/usr/bin/perl -l

use strict;
use warnings;

for (0, 1) {
   my $re = ($_ == 0 ? "" : "bar");
   my $x = "foo";
   $x =~ m/o/o;

   if ($x =~ /$re/) {
     print "$x =~ /$re/";
   }
}
__END__
The output of this:
   f o =~ //
   f o =~ /bar/


The reason for this is:
$re is empty the first time the loops run so the last succesfull  
pattern gets used.

That is m/o/o.
The /o modifier is not dropped so for the next iteration the loop  
still matches $x against m/o/  (because of the /o-modifier).


Currently in perlop:
If the PATTERN evaluates to the empty string, the last successfully  
matched regular expression is used instead. In this case, only the g  
and c flags on the empty pattern is honoured - the other flags are  
taken from the original pattern. If no match has previously succeeded,  
this will (silently) act instead as a genuine empty pattern (which  
will always match).


Should the docs be clearer about the empty pattern and the copying of  
the /o modifier? (which can be really confusing...)


Also,

Add: use re 'eval' to the code:

#!/usr/bin/perl -l

use strict;
use warnings;

use re 'eval';

for (0, 1) {
   my $re = ($_ == 0 ? "" : "bar");
   my $x = "foo";
   $x =~ m/o/o;

   if ($x =~ /$re/) {
     print "$x =~ /$re/";
   }
}
__END__
Output:
   foo =~ //


So how exactly does  use re 'eval'  influences the regex?



Kind regards,

Bram

• Empty pattern and /o by Bram
• Re: Empty pattern and /o by demerphq
• Re: Empty pattern and /o by Abigail
• Re: Empty pattern and /o by demerphq