throop

In such cases, I mostly resort to enumeration:

   /(0[1-9]
   |1[2-9]
   |2[3-9]
   |3[4-9]
   |4[5-9]
   |5[6-9]
   |6[7-9]
   |7[8-9]
   |8[9-9])/
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will only capture pairs where the first digit is lower than the second digit. I'm not sure that there is a smarter/nicer version of that though ...

It's usually best/simplest to just filter out the bad results.

my @x;
while (/(?=((\d)(\d)))/g) {
   push @x, $1 if $3 > $2;
}
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You could also do that nicely using a lookup.

my %ok;
for my $i (1..8) {
   for my $j ($i+1..9) {
      $ok{"$i$j"} = 1;
   }
}

my @x = grep $ok{$_}, /(?=(\d\d))/g;
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You could build a complex pattern.

use Regexp::List qw( );

my @ok;
for my $i (1..8) {
   for my $j ($i+1..9) {
      push @ok, "$i$j";
   }
}

my $re = Regexp::List->new()->list2re(@ok);
my @x = /(?=($re))/g;
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In can be done programmatically in the regexp as well.

my @x;
push @x, $1 while /(?=((\d)(\d)(?(?{ $3 <= $2 })(?!))))/g;
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(It's pretty crazy that I was able to write the last one without errors and without checking perlre, so you probably shouldn't use that one.)

All tested.

push @x, $1 while /(?=((\d)(\d)(?(?{ $3 <= $2 })(?!))))/g;

Bravo! That's the one I was looking for. I had a vague idea that it could be done within the regex, but I didn't know how to start.

Of course, I had to study pelre for a half an hour to understand what you'd done. But I learned 3 or 4 new things about Perl on the way. The neat thing is, I'd looked at those parts of perlre before. But I couldn't make heads or tails of them without your example.

Thanks!

throop

It does look a little bit hacky and I'm not sure you will find it explicitly discussed in either perlre or perlretut, but it is a kosher Perl regex expression and you need not worry about using it.