I don't quite understand the question. A regexp cannot change the string, it can only extract portions of the string.

If you mean an substutation and you can evaluate the substitution part (but why would you then be limited to evaluate inside a s///?), you can use the following simple technique.

s/(.*)/substr $1 . '0' x 13, 0, 13/se
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You just create a new string with all the zeros you could possibly need at the end, and then take the chars from the beginning.

my @codes = (
  '12345678',
  '1234567890123',
  '1234567890123456',
);

my $len = 13;

for my $var (@codes) {
  my $fixed = substr $var . '0' x $len, 0, $len;
  print "$fixed\n";
}

__END__
1234567800000
1234567890123
1234567890123
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lodin

s/\b(\d+)\b/sprintf "%013d", $1/eg;
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If not, please tell us exactly what you can use - a single regular expression only matches stuff, but can't change the source string.

Update: I missed that the 0's should actually be appended at the end. You can try to use lodin's solution inside the substitution part, though. Or to rescue my original solution:

s/\b(\d+)\b/reverse sprintf "%013d", scalar reverse $1/eg;
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this is not working by me.

This is not very helpful.

Would you care to elaborate? In what way does it not work? Wrong output? if yes, which? Or does it raise an error? if so, which?

Also with sprintf

$barcode = sprintf("%.13s%s", $barcode, "0"x(13-length($barcode)));
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update thanks lodin, I misunderstood the question and assumed the input was always 13 or less characters. Fixed now

As this is just a cumbersome way to do concatenation it does not fullfil the requirement to make $barcode shorter if it's over 13 chars long.

lodin

sprintf("<%.13s>",  "123" . "0" x 13)

Simplifying a little the idea from moritz you could try this:

Note that the single space between each pair of regex is important, as stated in the doc you posted.