Re: Backref in a regex quantifier

I can't tell you why the backref doesn't work in a quantifier... but you could use the "postponed regex" construct (??{ code }):

my ($n, $v) = "3 aaa " =~ /^(\d+?) ((??{".{$1}"})) /;

print "n=$n, v=$v\n";   # prints "n=3, v=aaa"
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Re^2: Backref in a regex quantifier by Anonymous Monk on Jul 09, 2007 at 19:58 UTC
a small tweak would be to use the `$^N` pre-defined variable that "contains whatever was matched by the most-recently closed group (submatch)". this makes the sub-expression less sensitive to its position in a possibly complex regex expression. `perl -wMstrict -e "my $re = qr/ ^ (\d+?) ( (??{ qq(.{$^N}) }) ) /x; /$re/, print qq($_ [$1] [$2] \n) for @ARGV" 43fffxxx 54321234 12xxxxxxxxxxx 12xxxxxxxxxxxx 43fffxxx [4] [3fff] 54321234 [5] [43212] 12xxxxxxxxxxx [1] [2] 12xxxxxxxxxxxx [1] [2]` [download]	[reply] [d/l] [select]

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Re^2: Backref in a regex quantifier
by Anonymous Monk on Jul 09, 2007 at 19:58 UTC

$^N

perl -wMstrict -e "my $re = qr/ ^ (\d+?) ( (??{ qq(.{$^N}) }) ) /x;
/$re/, print qq($_ [$1] [$2] \n) for @ARGV"
43fffxxx 54321234 12xxxxxxxxxxx 12xxxxxxxxxxxx
43fffxxx [4] [3fff]
54321234 [5] [43212]
12xxxxxxxxxxx [1] [2]
12xxxxxxxxxxxx [1] [2]
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