Re^3: Explain a regexp matched group result

[ig]

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.

[jdd]

Thank you very much. So I cannot get individuals without removing the repetition and do a while ()... i.e.

use strict;
use warnings FATAL => 'all';
 
my $string = "aacbbbcac";
 
my $re1 = qr/((a+)?(b+)?(c))/;
while ($string =~ /$re1/g) {
  foreach (0..$#-) {
    printf "Group %d: <%s>\n", $_, defined($-[$_]) ? substr($string, $
+-[$_], $+[$_] - $-[$_]) : '';
  }
  print "\n";
}

Bad news for me, but again mega thanks for your very clear answer. ps:

Group 0: <aac>
Group 1: <aac>
Group 2: <aa>
Group 3: <>
Group 4: <c>

Group 0: <bbbc>
Group 1: <bbbc>
Group 2: <>
Group 3: <bbb>
Group 4: <c>

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

[ig]

I think your conclusion is correct in general, but if you know the structure of the RE then there are workarounds to the way the capture groups work. Consider:

#!C:/strawberry/perl/bin/perl.exe
#
use strict;
use warnings;
 
my $string = "aacbbbcac";
 
my $re1 = qr/((a+)?(b+)?(c))*/;
#my $re1 = qr/((a*)(b*)(c))*/;
#my $re1 = qr/((a+)?(b*)(c))*/;
if ($string =~ $re1) {
  my $start = 0;
  my @something;
  foreach (0..$#-) {
    if(defined($-[$_])) {
        $start = $-[$_] if($-[$_] > $start);
        if($-[$_] >= $start) {
            printf "Group %d: <%s>\n", $_, substr($string, $-[$_], $+[
+$_] - $-[$_]);
            $something[$_] = substr($string, $-[$_], $+[$_] - $-[$_]);
        } else {
            printf "Group %d: <%s> - but ignore it because it is from 
+a previous iteration of the outer capture group\n", $_, substr($strin
+g, $-[$_], $+[$_] - $-[$_]);
            $something[$_] = '';
        }
    } else {
            printf "Group %d: hasn't matched yet\n", $_;
            $something[$_] = '';
    }
  }

  print "$1 = " . join('', @something[2..4]) . "\n";
}

Which produces

Group 0: <aacbbbcac>
Group 1: <ac>
Group 2: <a>
Group 3: <bbb> - but ignore it because it is from a previous iteration
+ of the outer capture group
Group 4: <c>
ac = ac

If you are trying to write something that handles arbitrary REs, this approach is unlikely to work.

[jdd]

Formidable!

How to say

• yes: I want to handle arbitraries REs
• but only REs that passed my grammar, so I know in advance the overall structure.
Complicated but perfectly doable.

The origin of my question is the comparison of perl's regexp with ECMAScript regexp, namely Note 3 of chapter 15.10.2.5 of ECMA-262 grammar spec (a perfectlty sensible question since ECMA regexp are are copy of perl5's).

I have done the AST of any ECMAScript source (c.f. MarpaX::Languages::ECMA::AST) so now I was wondering about the actions associated to the grammar.