How can a Perl regex re-use part of the previous match for the next match?

Question

I need some Perl regular expression help. The following snippet of code:

use strict; 
use warnings; 
my $str = "In this example, A plus B equals C, D plus E plus F equals G and H plus I plus J plus K equals L"; 
my $word = "plus"; 
my @results = ();
1 while $str =~ s/(.{2}\b$word\b.{2})/push(@results,"$1\n")/e;
print @results;

Produces the following output:

A plus B
D plus E
2 plus F
H plus I
4 plus J
5 plus K

What I want to see is this, where a character already matched can appear in a new match in a different context:

A plus B
D plus E
E plus F
H plus I
I plus J
J plus K

How do I change the regular expression to get this result? Thanks --- Dan

Answer 1

General advice: Don't use s/// when you want m//. Be specific in what you match.

The answer is pos:

#!/usr/bin/perl -l

use strict;
use warnings;

my $str = 'In this example, ' . 'A plus B equals C, ' .
          'D plus E plus F equals G ' .
          'and H plus I plus J plus K equals L';

my $word = "plus";

my @results;

while ( $str =~ /([A-Z] $word [A-Z])/g ) {
    push @results, $1;
    pos($str) -= 1;
}

print "'$_'" for @results;

Output:

C:\Temp> b
'A plus B'
'D plus E'
'E plus F'
'H plus I'
'I plus J'
'J plus K'

Answer 2

You can use a m//g instead of s/// and assign to the pos function to rewind the match location before the second term:

use strict;
use warnings;

my $str  = 'In this example, A plus B equals C, D plus E plus F equals G and H plus I plus J plus K equals L';
my $word = 'plus';
my @results;

while ($str =~ /(.{2}\b$word\b(.{2}))/g) {
    push @results, "$1\n";
    pos $str -= length $2;
}
print @results;