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Re^5: regexp question

by furry_marmot (Pilgrim)
on Oct 28, 2011 at 20:02 UTC ( #934495=note: print w/ replies, xml ) Need Help??


in reply to Re^4: regexp question
in thread regexp question

You're missing the whole point here. Square brackets are a character class. If I try to match on [a-z0-9], I'm specifying one character that falls in the class of characters from a-z and 0-9. That is, I'm trying to match one character that could be any of those in the class.

But if I try to match on [.], I'm specifying one character that falls in the class of characters that are a period. In other words, [a-z] could match 'a', or 'b', or 'c', etc., but [.] can only ever match '.'. So [.] is exactly equal to '.' Thus it's a useless use of a character class.

To use another example of yours, [\d\s] will match one character that is either a digit or a space character. It could match 9, or 8, or ' '. \d and \s retain their "magic" even in a character class. [\d] = \d = [0-9]

The lesson here is don't use single-character classes.

--marmot


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Re^6: regexp question
by choroba (Abbot) on Oct 28, 2011 at 21:10 UTC
    [\d] = \d = [0-9]
    Not exactly true. See perlrecharclass:
    "\d" matches a single character that is considered to be a digit. What is considered a digit depends on the internal encoding of the source string and the locale that is in effect. If the source string is in UTF-8 format, "\d" not only matches the digits '0' - '9', but also Arabic, Devanagari and digits from other languages. Otherwise, if there is a locale in effect, it will match whatever characters the locale considers digits. Without a locale, "\d" matches the digits '0' to '9'. See "Locale, EBCDIC, Unicode and UTF-8".

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