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Re^5: Explain a regexp matched group result

by ig (Vicar)
on Oct 28, 2013 at 19:31 UTC ( #1060050=note: print w/replies, xml ) Need Help??

in reply to Re^4: Explain a regexp matched group result
in thread Explain a regexp matched group result

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.

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Re^6: Explain a regexp matched group result
by jdd (Acolyte) on Oct 28, 2013 at 19:45 UTC

    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 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.

      The note is interesting: they are highlighting this difference between ECMA script REs and Perl REs.

      The RE (x+)? is very similar to (x*), except that the latter will always match (and, therefore, never have the value from a previous match if it is in an enclosing repeating group. This is similar to the requirement in Note 3: "Step 4 of the RepeatMatcher clears Atom's captures each time Atom is repeated." Because it always matches it always has a value from the last repeat of the outer repeating group, as if it was reset for each repeat, except that the value is '' instead of undef in the case that x did not match. This is an easy transformation.

      I appreciate that you don't want to change the RE but you say you are parsing it, so perhaps you can make some systematic transformations.


      use strict; use warnings; use Data::Dumper::Concise; my $string = "aacbbbcac"; my $re = '((a+)?(b+)?(c))*'; # transform '(x+)?' to '(x*)' assuming 'x' is monolithic $re =~ s/\Q+)?/*)/g; print "re = $re\n"; my $re1 = qr/$re/; if ($string =~ $re1) { my @something; foreach (0..$#-) { if(defined($-[$_])) { my $substring = substr($string, $-[$_], $+[$_] - $-[$_]); # ${$_} also works, except where $_ = 0 no strict 'refs'; print "\$substring = $substring = ${$_}\n"; # transform '' to undef $substring = undef if($substring eq ''); # assert: $substring is now as specified by # Standard ECMA-262, 5.1 Edition / June 2011 # Section Note 3 printf "Group %d: <%s>\n", $_, $substring // ''; $something[$_] = $substring; } } print Dumper(\@something); }


      re = ((a*)(b*)(c))* $substring = aacbbbcac = Group 0: <aacbbbcac> $substring = ac = ac Group 1: <ac> $substring = a = a Group 2: <a> $substring = = Group 3: <> $substring = c = c Group 4: <c> [ "aacbbbcac", "ac", "a", undef, "c" ]

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