It is possible using the postponed regular subexpression. 
/([0-8])[a-z](??{ $1 + 1 })/

[download]

See perlre for details.

Updated: Simplified, don't use named capture groups.

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

use Test2::V0 '-no_srand' => 1;

my @cases = (
    ['13s4', '3s4'],
    ['3s44', '3s4'],
    ['0a1',  '0a1'],
    ['0a2',  undef],
    ['9y10', undef]
);

for my $case (@cases) {
    my ($line, $expected) = @$case;
    my $result;
    $result = $& if $line =~ /([0-8])[a-z](??{ $1 + 1 })/;
    is $result, $expected, $line;
}

done_testing;

[download]

map{substr$_->[0],$_->[1]||0,1}[\*||{},3],[[]],[ref qr-1,-,-1],[{}],[sub{}^*ARGV,3]

In reply to Re: arithmetic in a regex ? by choroba
in thread arithmetic in a regex ? by Anonymous Monk

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