Re: Regular expressions: interpolate the variable in the value of the number of repetitions for the group

[kcott]

G'day 0day,

The perlre documentation, in the Extended Patterns section, says this about the (??{ code }) construct:

"During the matching of this sub-pattern, it has its own set of captures which are valid during the sub-match, but are discarded once control returns to the main pattern."

So, while /(\d)((??{'.*?\n' x $1}))/ captures $2 as you state, attempting to capture sub-patterns within (??{ code }) to $3, $4, etc. won't work. Furthermore, unless your target string always begins with '3' you won't know how many $<integer> variables will be available.

The following script achieves the result you're after by storing the sub-pattern captures in an array.

#!/usr/bin/env perl -T

use 5.010;
use strict;
use warnings;
use re qw{taint eval};

my $x = "3aaaa\nbbbb\nccccc\nddddd";

my @captures;

$x =~ /\A (\d+) ( (??{ '(.+?(?:\n|$))(?{ push @captures, $^N })' x $1 
+}) )/mx;

say '[1]', $1, '[1]';
say '[2]', $2, '[2]';
say '[', $_ + 3, ']', $captures[$_], '[', $_ + 3, ']' for 0 .. $#captu
+res;

Here's the output which, as you can see, is somewhat fudged to give a sense of what $3, $4, etc. would have been:

$ pm_1047667_regex.pl
[1]3[1]
[2]aaaa
bbbb
ccccc
[2]
[3]aaaa
[3]
[4]bbbb
[4]
[5]ccccc
[5]

Changing the first character of the target string to '2':

$ pm_1047667_regex.pl
[1]2[1]
[2]aaaa
bbbb
[2]
[3]aaaa
[3]
[4]bbbb
[4]

And to '4':

$ pm_1047667_regex.pl
[1]4[1]
[2]aaaa
bbbb
ccccc
ddddd[2]
[3]aaaa
[3]
[4]bbbb
[4]
[5]ccccc
[5]
[6]ddddd[6]

Notes:

• Read the security concerns regarding (?{ code }) and (??{ code }) mentioned in both the re (pragma) and perlre documentation. The script I've posted uses both the -T switch and use re 'taint': this is more to highlight the issue rather than any real worries about matching parts of "3aaaa\nbbbb\nccccc\nddddd".
• Also read the experimental warnings for both (?{ code }) and (??{ code }) in perlre. I'm using 5.18.0, you may get different results with another version.

[0day]

Wow...
Thank you very much kcott. Thank that you explained, your decision is the most elegant.

Many thanks to all who tried to help.