Re: get next higher number

[AnomalousMonk]

Can I do it some how with regex?

As hippo has alluded, a regex approach would seem far from ideal. Howsoever, the following seems to work, but requires Perl version 5.10+ for the  (?(condition)yes-pattern|no-pattern) construct (which embeds the experimental  (?{ CODE }) feature; see Extended Patterns in perlre):

c:\@Work\Perl>perl -wMstrict -le
"use 5.010;
 ;;
 use Test::More 'no_plan';
 use Test::NoWarnings;
 ;;
 my @values = qw(
   11673326 11673329 11673325 11673330 11673321 11673335
   );
 ;;
 my $sorted_str = join ' ', sort { $a <=> $b } @values;
 ;;
 sub next_higher {
   local our ($from) = @_;
   ;;
   return $1 if $sorted_str =~ m{
     \b (\d+) \b
     (?(?{ $from >= $^N }) (*FAIL))
     }xms;
   ;;
   return;
   }
 ;;
 is next_higher(11673326), 11673329;
 is next_higher(11673335), undef;
 is next_higher(11673321), 11673325;
 done_testing;
"
ok 1
ok 2
ok 3
1..3
ok 4 - no warnings
1..4

I leave further testing to you :)

(NB: From, I believe, Perl version 5.18 on, the
    local our ($from) = @_;
statement in the  next_higher() function can be replaced with a more familiar lexical
    my ($from) = @_;
statement and the regex will work properly; can't test this ATM.)

Update: When discussing Perl version dependency in the first paragraph, I should also have mentioned that  (*FAIL) was introduced with 5.10 (see Special Backtracking Control Verbs in perlre). However, the effect of  (*FAIL) is exactly duplicated by  (?!) in pre-5.10 regexen, so  (?(condition)yes-pattern|no-pattern) remains the critical problem.