http://qs321.pair.com?node_id=731328


in reply to Extracting the number of repetitions from a regex

It's a scoping issue. Apparently, the regex uses the variables that existed when it was first compiled, while you have declared a new set of variables on every pass. Or something like that. Anyway, this works: 
my @strings = qw( aaabbbb ab abb aabb aaabb aabbb );

{
  my ($a_counter, $b_counter);
  for my $string ( @strings )
  {
    $a_counter = 0;
    $b_counter = 0;
    print "In $string there were $a_counter 'a's and $b_counter 'b's. 
+\n" if ( $string =~ /(a(?{$a_counter ++;}))+(b(?{$b_counter ++;}))+/ 
+);
  }
}

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Re^2: Extracting the number of repetitions from a regex
by pat_mc (Pilgrim) on Dec 18, 2008 at 20:30 UTC
    Roy -

    Thanks for shedding some light onto this issue. I am grateful for your functional modification of the code which does indeed produce the desired output.

    As for the reasoning why the original code behaves the way it does, however, I am not sure I follow your explanation. Since the regex is missing the //o modifier I would have expected it to get re-compiled on every iteration of the for-loop (please correct me if I am wrong here!). If it doesn't - why not?

    On the other hand, your reasoning is supported by the output of the following code modification: 
    my @strings = qw( aaabbbb ab abb aabb aaabb aabbb );

my $a_counter;
my $b_counter;

for my $string ( @strings )
{
    print "In $string there were $a_counter 'a's and $b_counter 'b's.\
+n" if ( $string =~ /(a(?{$a_counter ++; }  ))+(b(?{$b_counter ++;}))+
+/ );
}

    [download]
    which produces 
    In aaabbbb there were 3 'a's and 4 'b's.
In ab there were 4 'a's and 5 'b's.
In abb there were 5 'a's and 7 'b's.
In aabb there were 7 'a's and 9 'b's.
In aaabb there were 10 'a's and 11 'b's.
In aabbb there were 12 'a's and 14 'b's.

    [download]
    So: Can anybody please explain conclusively what is happening here in the compilation of the regex?
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