Re: Arbitrary number of captures in a regular expression

[bsb]

in reply to Arbitrary number of captures in a regular expression

It's not fancy or general, but it's simple and may be adaptable to your purpose:

$ms = '(?: m (\d+))?' x 5; # 1)
print join(",", /^foo$ms bar$/ ),"\n" for (
    'foo m 1 m 2 m 3 m 4 bar',
    'foo m 2 m 4 m 7 bar',
    'foo m 1 bar',
);
__END__
1,2,3,4,
2,4,7,,
1,,,,

1) Where "any number" is equal to 5...