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Asked: 2026-06-15T00:54:37+00:00

I want to understand how can I do arithmetic on matched sub-patterns in perl

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I want to understand how can I do arithmetic on matched sub-patterns in perl regex.
This is just a sample code and I want to understand how can I use \1 (already matched sub-pattern. in this case – 7) to match pattern+1 (8)

my $y = 77668;
if($y =~ /(\d)\1(\d)\2\1+1/)   #How to increment a previously
                               #matched sub-pattern and form a pattern?
{
    print $y;
}

EDIT

From the answers, I see that pattern arithmetic is not possible.

This is what I want to achieve.
I want to form a regex which will match this pattern:

N-3N-2N-1NNN+1N+2N+3    (N = 3,4,5,6
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1 Answer

  1. Editorial Team

    Its possible via regex code blocks:

    my $y = 77668;
if($y =~ /(\d)\1(\d)\2(??{$1+1})/ ) {
    print $y;
}

    In this snippet (??{ CODE }) returns another regex that must match, so this regex looks like “8” ($1+1). As a result, whole regex will match only if 5th digit is greather and 1st by 1. But drawback with 1st digit is 9, this code block will return “10”, so possible its wrong behavior, but you said nothing about what must be done in this case.

    Now about N-3N-2N-1NNN+1N+2N+3 question, you can match it with this regex:

    my $n = 5;
if( $y =~ /(??{ ($n-3).($n-2).($n-1).$n.($n+1).($n+2).($n+3) })/ ){

    Or more “scalable” way:

    my $n = 5;
if( $y =~ /(??{ $s=''; $s .= $n+$_ foreach(-3..3); $s; })/ ){

    Again, what we must do if $n == 2 ?? $n-3 will be -1. Its not a simply digit cus it have sign, so you should think about this cases.

    One another way. Match what we have and then check it.

    if( $y =~ /(\d)(\d)(\d)(\d)(\d)(\d)(\d)/ ) {
    if( $1 == ($4-3) && $2 == ($4-2) && $3 == ($4-1) && $6 == ($4+1) && $7 == ($4+2) && $7 == ($4+3) ) {
        #...

    Seems this method litle bit clumsy, but its obivious to everyone (i hope).

    Also, you can optimize your regex since 7 ascending digits streak is not so frequent combination, plus get some lulz from co-workers xD:

    sub check_number {
    my $i;
    for($i=1; $i<length($^N); $i++) {
        last if substr($^N, $i, 1)<=substr($^N, $i-1, 1);
    }
    return $i<length($^N) ? "(*FAIL)" : "(*ACCEPT)";
}

if( $y =~ /[0123][1234][2345][3456][4567][5678][6789](??{ check_number() })/ ) {

    Or… Maybe most human-friendly method:

    if( $y =~ /0123456|1234567|2345678|3456789/ ) {

    Seems last variant is bingo xD Its good example about not searching regex when things are so simple)

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