=over 4
=item *
+
\d is a digit and represents [0-9]
=item *
+
\s is a whitespace character and represents [\ \t\r\n\f]
=item *
+
\w is a word character (alphanumeric or _) and represents [0-9a-zA-Z_]
=item *
+
\D is a negated \d; it represents any character but a digit [^0-9]
=item *
+
\S is a negated \s; it represents any non-whitespace character [^\s]
=item *
+
\W is a negated \w; it represents any non-word character [^\w]
=item *
+
The period '.' matches any character but "\n"
=back
=over 4
=item *
+
no modifiers (//): Default behavior. C<'.'> matches any character
except C<"\n">. C<^> matches only at the beginning of the string and
C<$> matches only at the end or before a newline at the end.
=item *
+
s modifier (//s): Treat string as a single long line. C<'.'> matches
any character, even C<"\n">. C<^> matches only at the beginning of
the string and C<$> matches only at the end or before a newline at the
end.
=item *
+
m modifier (//m): Treat string as a set of multiple lines. C<'.'>
matches any character except C<"\n">. C<^> and C<$> are able to match
at the start or end of I<any> line within the string.
=item *
+
both s and m modifiers (//sm): Treat string as a single long line, but
detect multiple lines. C<'.'> matches any character, even
C<"\n">. C<^> and C<$>, however, are able to match at the start or end
=over 4
-=item 0 Start with the first letter in the string 'a'.
+=item 0
+
+Start with the first letter in the string 'a'.
+
+=item 1
+
+Try the first alternative in the first group 'abd'.
+
+=item 2
-=item 1 Try the first alternative in the first group 'abd'.
+Match 'a' followed by 'b'. So far so good.
-=item 2 Match 'a' followed by 'b'. So far so good.
+=item 3
-=item 3 'd' in the regexp doesn't match 'c' in the string - a dead
+'d' in the regexp doesn't match 'c' in the string - a dead
end. So backtrack two characters and pick the second alternative in
the first group 'abc'.
-=item 4 Match 'a' followed by 'b' followed by 'c'. We are on a roll
+=item 4
+
+Match 'a' followed by 'b' followed by 'c'. We are on a roll
and have satisfied the first group. Set $1 to 'abc'.
-=item 5 Move on to the second group and pick the first alternative
+=item 5
+
+Move on to the second group and pick the first alternative
'df'.
-=item 6 Match the 'd'.
+=item 6
+
+Match the 'd'.
+
+=item 7
-=item 7 'f' in the regexp doesn't match 'e' in the string, so a dead
+'f' in the regexp doesn't match 'e' in the string, so a dead
end. Backtrack one character and pick the second alternative in the
second group 'd'.
-=item 8 'd' matches. The second grouping is satisfied, so set $2 to
+=item 8
+
+'d' matches. The second grouping is satisfied, so set $2 to
'd'.
-=item 9 We are at the end of the regexp, so we are done! We have
+=item 9
+
+We are at the end of the regexp, so we are done! We have
matched 'abcd' out of the string "abcde".
=back
=over 4
-=item * C<a?> = match 'a' 1 or 0 times
+=item *
+
+C<a?> = match 'a' 1 or 0 times
+
+=item *
+
+C<a*> = match 'a' 0 or more times, i.e., any number of times
-=item * C<a*> = match 'a' 0 or more times, i.e., any number of times
+=item *
-=item * C<a+> = match 'a' 1 or more times, i.e., at least once
+C<a+> = match 'a' 1 or more times, i.e., at least once
+
+=item *
-=item * C<a{n,m}> = match at least C<n> times, but not more than C<m>
+C<a{n,m}> = match at least C<n> times, but not more than C<m>
times.
-=item * C<a{n,}> = match at least C<n> or more times
+=item *
+
+C<a{n,}> = match at least C<n> or more times
-=item * C<a{n}> = match exactly C<n> times
+=item *
+
+C<a{n}> = match exactly C<n> times
=back
=over 4
=item *
+
Principle 0: Taken as a whole, any regexp will be matched at the
earliest possible position in the string.
=item *
+
Principle 1: In an alternation C<a|b|c...>, the leftmost alternative
that allows a match for the whole regexp will be the one used.
=item *
+
Principle 2: The maximal matching quantifiers C<?>, C<*>, C<+> and
C<{n,m}> will in general match as much of the string as possible while
still allowing the whole regexp to match.
=item *
+
Principle 3: If there are two or more elements in a regexp, the
leftmost greedy quantifier, if any, will match as much of the string
as possible while still allowing the whole regexp to match. The next
=over 4
-=item * C<a??> = match 'a' 0 or 1 times. Try 0 first, then 1.
+=item *
+
+C<a??> = match 'a' 0 or 1 times. Try 0 first, then 1.
+
+=item *
-=item * C<a*?> = match 'a' 0 or more times, i.e., any number of times,
+C<a*?> = match 'a' 0 or more times, i.e., any number of times,
but as few times as possible
-=item * C<a+?> = match 'a' 1 or more times, i.e., at least once, but
+=item *
+
+C<a+?> = match 'a' 1 or more times, i.e., at least once, but
as few times as possible
-=item * C<a{n,m}?> = match at least C<n> times, not more than C<m>
+=item *
+
+C<a{n,m}?> = match at least C<n> times, not more than C<m>
times, as few times as possible
-=item * C<a{n,}?> = match at least C<n> times, but as few times as
+=item *
+
+C<a{n,}?> = match at least C<n> times, but as few times as
possible
-=item * C<a{n}?> = match exactly C<n> times. Because we match exactly
+=item *
+
+C<a{n}?> = match exactly C<n> times. Because we match exactly
C<n> times, C<a{n}?> is equivalent to C<a{n}> and is just there for
notational consistency.
=over 4
=item *
+
Principle 3: If there are two or more elements in a regexp, the
leftmost greedy (non-greedy) quantifier, if any, will match as much
(little) of the string as possible while still allowing the whole
=over 4
-=item 0 Start with the first letter in the string 't'.
+=item 0
+
+Start with the first letter in the string 't'.
-=item 1 The first quantifier '.*' starts out by matching the whole
+=item 1
+
+The first quantifier '.*' starts out by matching the whole
string 'the cat in the hat'.
-=item 2 'a' in the regexp element 'at' doesn't match the end of the
+=item 2
+
+'a' in the regexp element 'at' doesn't match the end of the
string. Backtrack one character.
-=item 3 'a' in the regexp element 'at' still doesn't match the last
+=item 3
+
+'a' in the regexp element 'at' still doesn't match the last
letter of the string 't', so backtrack one more character.
-=item 4 Now we can match the 'a' and the 't'.
+=item 4
+
+Now we can match the 'a' and the 't'.
-=item 5 Move on to the third element '.*'. Since we are at the end of
+=item 5
+
+Move on to the third element '.*'. Since we are at the end of
the string and '.*' can match 0 times, assign it the empty string.
-=item 6 We are done!
+=item 6
+
+We are done!
=back
=over 4
-=item * specifying the task in detail,
+=item *
+
+specifying the task in detail,
+
+=item *
+
+breaking down the problem into smaller parts,
-=item * breaking down the problem into smaller parts,
+=item *
+
+translating the small parts into regexps,
-=item * translating the small parts into regexps,
+=item *
-=item * combining the regexps,
+combining the regexps,
+
+=item *
-=item * and optimizing the final combined regexp.
+and optimizing the final combined regexp.
=back
# prints 'Hi Mom!'
$x =~ /aaa(?{print "Hi Mom!";})def/; # doesn't match,
# no 'Hi Mom!'
+
+Pay careful attention to the next example:
+
$x =~ /abc(?{print "Hi Mom!";})ddd/; # doesn't match,
# no 'Hi Mom!'
+ # but why not?
+
+At first glance, you'd think that it shouldn't print, because obviously
+the C<ddd> isn't going to match the target string. But look at this
+example:
+
+ $x =~ /abc(?{print "Hi Mom!";})[d]dd/; # doesn't match,
+ # but _does_ print
+
+Hmm. What happened here? If you've been following along, you know that
+the above pattern should be effectively the same as the last one --
+enclosing the d in a character class isn't going to change what it
+matches. So why does the first not print while the second one does?
+
+The answer lies in the optimizations the REx engine makes. In the first
+case, all the engine sees are plain old characters (aside from the
+C<?{}> construct). It's smart enough to realize that the string 'ddd'
+doesn't occur in our target string before actually running the pattern
+through. But in the second case, we've tricked it into thinking that our
+pattern is more complicated than it is. It takes a look, sees our
+character class, and decides that it will have to actually run the
+pattern to determine whether or not it matches, and in the process of
+running it hits the print statement before it discovers that we don't
+have a match.
+
+To take a closer look at how the engine does optimizations, see the
+section L<"Pragmas and debugging"> below.
+
+More fun with C<?{}>:
+
$x =~ /(?{print "Hi Mom!";})/; # matches,
# prints 'Hi Mom!'
$x =~ /(?{$c = 1;})(?{print "$c";})/; # matches,
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