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Scott Klement wrote:
pattern = '(\S\S{40,}|.{1,40})(\s+|$)' + x'00';
The next element of the array matches your first subexpression.
Subexpressions are the things in parenthesis. Your first one is
(\S\S{40,}|.{1,40}) -- which means 40 or more of the letter S or 1-40 of
any character.
Not to put words in Aaron's mouth, but I suspect that when he's using \s
and \S he's looking for whitespace and non-whitespace respectively. Do
the C regex APIs support Extended Regular Expression syntax,
specifically the shorthand character classes? [1]
If I'm right about Aaron's intention with \s and \S, then I'm a bit
confused about the '\S\S{40,} part. I think that if the search string
contained 41 or more consecutive non-whitespace characters, that would
be what was matched. I haven't tested this.
HTH,
Adam
[1]
http://en.wikipedia.org/wiki/Regular_expression#POSIX_Extended_Regular_Expressions
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