This is the best I could come up with.. It's possible that this could be improved, as it's a little messy. Anyway, in P::RD you can pass arguments to a subrule using rule[args], and within the rule use the @arg or %arg variable to fetch those args. 
#!/usr/bin/perl
use strict;
use warnings;
use Test::More tests => 6;
use Parse::RecDescent;

my $p = Parse::RecDescent->new(<<'END_GRAMMAR') or die;
        rec: int elem[ num => $item[1] ] /\Z/
             { $item[2] }

        int: /\d+\b/

 ## match either the elem_many or the elem_one rule, and pass
 ## the number along to it..

       elem: { $arg{num} > 1 ? "many" : "one" }
             <matchrule:elem_$item[1]>[ num => $arg{num} ]
             { $item[2] }

  elem_many: elem_one elem[ num => $arg{num}-1 ]
             { [ $item[1], @{$item[2]} ] }

   elem_one: /\S+(?!\S)/
             { [$item[1]] }

END_GRAMMAR

# ok( $p->rec('0'));
ok( $p->rec('1 foo'));
ok( $p->rec('2 foo bar'));
ok(!$p->rec('1'));
ok(!$p->rec('1 foo bar'));
ok( $p->rec('3 foo bar baz'));
ok(!$p->rec('3 foo bar baz flab'));

[download]
It works for everything except the zero case, which shouldn't be too bad to add as a degenerate case.

Also note that I added a /\Z/ token to the main rule, so we would be assured that the rule was matching the whole string. And I also changed /\S+/ to /\S+(?!\S)/ to make sure each elem was a maximum-length word.

BTW, you could do this simple problem with extended regexes: something like /(\d+)\s+(??{ "(?:\\S+(?>\\S)\\s*){$1}" })/ off the top of my head. But if elem_one were matching anything significantly more complicated /\S+/, you'd have to do something like this in a grammar.

blokhead

In reply to Re: Parse::RecDescent and Dynamically Matched Subrule Repetition by blokhead
in thread Parse::RecDescent and Dynamically Matched Subrule Repetition by lhoward

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