Can you salvage my negative lookbehind example for commifying numbers?

Question

In the \"Advanced Regular Expresssion\" chapter in Mastering Perl, I have a broken example for which I can\'t figure out a nice fix. The example is perhaps trying to be too clev

Answer 1

I don't think it's possible without some form of variable-width look-behind. The addition of the \K assertion in 5.10 provides a way of faking variable-width positive look-behind. What we really need is variable-width negative look-behind but with a little creativity and a lot of ugliness we can make it work:

use 5.010;
$_ = '$1234567890.123456789';
s/(?<!\.)(?:\b|\G)\d+?\K(?=(?:\d\d\d)+\b)/,/g;
say;  # $1,234,567,890.123456789

If there was ever a pattern that begged for the /x notation it's this one:

s/
  (?<!\.)        # Negative look-behind assertion; we don't want to match
                 # digits that come after the decimal point.

  (?:            # Begin a non-capturing group; the contents anchor the \d
                 # which follows so that the assertion above is applied at
                 # the correct position.

    \b           # Either a word boundary (the beginning of the number)...

    |            # or (because \b won't match at subsequent positions where
                 # a comma should go)...

    \G           # the position where the previous match left off.

  )              # End anchor grouping

  \d+?           # One or more digits, non-greedily so the match proceeds
                 # from left to right. A greedy match would proceed from
                 # right to left, the \G above wouldn't work, and only the
                 # rightmost comma would get placed.

  \K             # Keep the preceding stuff; used to fake variable-width
                 # look-behind

                 # <- This is what we match! (i.e. a position, no text)

  (?=            # Begin a positive look-ahead assertion

    (?:\d\d\d)+  # A multiple of three digits (3, 6, 9, etc.)

    \b           # A word (digit) boundary to anchor the triples at the
                 # end of the number.

  )              # End positive look-ahead assertion.
/,/xg;

Answer 2

If you have to post on Stack Overflow asking if somebody can figure out how to do this with negative lookbehind, then it's obviously not a good example of negative lookbehind. You'd be better off thinking up a new example rather than trying to salvage this one.

In that spirit, how about an automatic spelling corrector?

s/(?<![Cc])ei/ie/g; # Put I before E except after C

(Obviously, that's not a hard and fast rule in English, but I think it's a more realistic application of negative lookbehind.)

Answer 3

I don't think this is what you are after (especially becaue the negative look-behind assertion has been dropped), but I guess, your only option is to slurp up the decimal places like in this example:

s/
  (?:
    (?<=\d)
    (?=(?:\d\d\d)+\b)
   |
    ( \d{0,3} \. \d+ )
  )
 / $1 ? $1 : ',' /exg;

P.S. I think it is a good example when not used as the first one in the book, as it demonstrates some of the pitfalls and limitations of look-around assertions.