Dear Monks, I have a weird problem that I don't know how to solve. But it sounds like a very-very simple problem.

I have a very long string which has a list of big integers. And then I have a pointer which points to one of the digits of one of the integers. As long as the pointer points to the first digit of an integer, I can extract it using regex, but what if the pointer happens to point to the 3rd or 4th digit of the integer? Then I don't know how to capture the whole integer. Is there a way to capture backwards and forwards at the same time? 


#!/usr/bin/perl

use strict;
use warnings;

my $S = 'Hello World 123456 !!! 789 ...';
my $P = 14;   # $P points to digit '3'

# I want to capture 123456 but I can only capture 3456 with this regex
+:

my $INT = ($S =~ m/^.{$P}(\d+)/) ? $1 : '';

print "\n :$INT: \n";

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In reply to How to capture backwards using regex? by harangzsolt33

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