Edumacation (I hope, anyway) by example :-)

#!/usr/bin/perl
use 5.016;
use warnings;

# 1060027
 
my $string = "aacbbbcac";
 
my $re1 = qr/((a+)?(b+)?(c))*/;
my $re2 = qr/(a.*?)(b.+?)(c)(.*)/;  # ADDED ON SUSPICION THAT OP'S REG
+EX FU 
                                    # MIGHT BE INCREMENTED BY THIS EXA
+MPLE

if ($string =~ $re1) {
  say "Here are the captures: $1, $2, $3)\n";  # ADDED line
  foreach (0..$#-) {
    printf "Group %d: <%s>\n", $_, substr($string, $-[$_], $+[$_] - $-
+[$_]);
  }
}
say "\n";

if ( $string =~ $re2 ) {
    say "USING \$re2, here are the captures \$1..\$4: $1, $2, $3, $4)\
+n";

    foreach (0..$#-) {
    printf "Group %d: <%s>\n", $_, substr($string, $-[$_], $+[$_] - $-
+[$_]);
  }
}


=head   #ONLY the "ADDED" lines are changed from OP's original

Here are the captures: ac, a, bbb)

Group 0: <aacbbbcac>
Group 1: <ac>
Group 2: <a>
Group 3: <bbb>
Group 4: <c>


USING $re2, here are the captures $1..$4: aac, bbb, c, ac)

Group 0: <aacbbbcac>
Group 1: <aac>
Group 2: <bbb>
Group 3: <c>
Group 4: <ac>

=cut
[download]

Thank you very much to you too, now I understand -;

Could you please explain what exactly you expect to get as a result from your example string?

Whatever the values, I was expecting the insides $2, $3 and $4 to be equivalent to $1 after concatenation. Why is it not the case?

From perlvar:

$-[0] is the offset of the start of the last successful match.

You have my $re1 = qr/((a+)?(b+)?(c))*/;

Your outer capture group may be repeated zero or more times. In the case of your test string, "aacbbbcac", it matches three times:

At the first repeat, it matches "aac" with $1 being "aac", $2 being "aa", $3 being undef (not matched) and $4 being "c", but because of the repeat count it doesn't stop there, so you never see these values.

At the second repeat $1 is "bbbc", $2 remains "aa" (group (a+) didn't match in this repeat but $2 is the 'last successful match'), $3 is "bbb" and $4 is "c", but it doesn't stop there, so you don't see these values either.

The third and final repeat sets $1 to "ac", $2 to "a" leaves $3 as it was at the last successful match (i.e. "bbb") and sets $4 to "c".

So, the issue is that the capture groups return the last successful match rather than the last match or failure as the case may be.