Errata

The Perl replacement text appears as
\$%\*\$1\\1

This wil not preserve the backlash before the asterisk in the original text:
$%\*$1\1

The replacement text should have two backslashes before the asterisk:
\$%\\*\$1\\1

Ruby 1.9 regular expression editor (rubular.com) shows the following error in this code: Forward slashes must be escaped

Current text:

In Java, JavaScript, PCRE, and Ruby, ?\w? is always identical to ?[a-zA-Z0-9_]?.


However, Ruby 1.9 regular expressions editor (rubular) highlights for \w special characters of European languages (such as ?, ă, ?) and Cyrillic characters as well.

The last paragraph says "For variable repetition, we use the quantifier <{n,m}>, where n is a POSITIVE NUMBER and m in greater than n....". I think this should read "where n is a NON-NEGATIVE INTEGER etc....", being as n can be 0. Saying that it is an integer (rather than a number) also makes it explicit that it has to be an integer, although I'm sure most people would come to this conclusion regardless of whether the term integer or number is used.

On Page 250, you explained that a trailing line break adds an extra empty line, then how do you deal with a leading line break?

If you type "\nabc" in a text editor, you will get TWO lines instead of ONE. Thus I would argue that "\nabc\nabc\nabc\nabc\nabc" also contains SIX lines with an empty first line, but it will be accepted by the given regexes.

Am I getting something wrong here? Thanks.

Note from the Author or Editor:
You've gotten it right. You can fix the handling for strings starting with a line break by using \A(?>[^\r\n]*(?>\r\n?|\n)?){0,4}[^\r\n]*\z or equivalent. The necessary changes to the recipe are extensive due to the numerous variations listed.