In addition, Perl defines the following:
\w Match a "word" character (alphanumeric plus "_")
- \W Match a non-word character
+ \W Match a non-"word" character
\s Match a whitespace character
\S Match a non-whitespace character
\d Match a digit character
equivalent to C<(?:\PM\pM*)>
\C Match a single C char (octet) even under utf8.
-A C<\w> matches a single alphanumeric character, not a whole word.
+A C<\w> matches a single alphanumeric character or C<_>, not a whole word.
Use C<\w+> to match a string of Perl-identifier characters (which isn't
the same as matching an English word). If C<use locale> is in effect, the
list of alphabetic characters generated by C<\w> is taken from the
of your string, see the previous reference. The actual location
where C<\G> will match can also be influenced by using C<pos()> as
an lvalue. See L<perlfunc/pos>.
-
+
The bracketing construct C<( ... )> creates capture buffers. To
-refer to the digit'th buffer use \E<lt>digitE<gt> within the
+refer to the digit'th buffer use \<digit> within the
match. Outside the match use "$" instead of "\". (The
-\E<lt>digitE<gt> notation works in certain circumstances outside
+\<digit> notation works in certain circumstances outside
the match. See the warning below about \1 vs $1 for details.)
Referring back to another part of the match is called a
I<backreference>.
if (/(.)\1/) { # find first doubled char
print "'$1' is the first doubled character\n";
}
-
+
if (/Time: (..):(..):(..)/) { # parse out values
$hours = $1;
$minutes = $2;
$seconds = $3;
}
-
+
Several special variables also refer back to portions of the previous
match. C<$+> returns whatever the last bracket match matched.
C<$&> returns the entire matched string. (At one point C<$0> did
Backslashed metacharacters in Perl are alphanumeric, such as C<\b>,
C<\w>, C<\n>. Unlike some other regular expression languages, there
are no backslashed symbols that aren't alphanumeric. So anything
-that looks like \\, \(, \), \E<lt>, \E<gt>, \{, or \} is always
+that looks like \\, \(, \), \<, \>, \{, or \} is always
interpreted as a literal character, not a metacharacter. This was
once used in a common idiom to disable or quote the special meanings
of regular expression metacharacters in a string that you want to
-use for a pattern. Simply quote all non-alphanumeric characters:
+use for a pattern. Simply quote all non-"word" characters:
$pattern =~ s/(\W)/\\$1/g;
For look-behind see below.
-=item C<(?E<lt>=pattern)>
+=item C<(?<=pattern)>
-A zero-width positive look-behind assertion. For example, C</(?E<lt>=\t)\w+/>
+A zero-width positive look-behind assertion. For example, C</(?<=\t)\w+/>
matches a word that follows a tab, without including the tab in C<$&>.
Works only for fixed-width look-behind.
Better yet, use the carefully constrained evaluation within a Safe
module. See L<perlsec> for details about both these mechanisms.
-=item C<(?p{ code })>
+=item C<(??{ code })>
B<WARNING>: This extended regular expression feature is considered
highly experimental, and may be changed or deleted without notice.
(?:
(?> [^()]+ ) # Non-parens without backtracking
|
- (?p{ $re }) # Group with matching parens
+ (??{ $re }) # Group with matching parens
)*
\)
}x;
-=item C<(?E<gt>pattern)>
+=item C<< (?>pattern) >>
B<WARNING>: This extended regular expression feature is considered
highly experimental, and may be changed or deleted without notice.
It may also be useful in places where the "grab all you can, and do not
give anything back" semantic is desirable.
-For example: C<^(?E<gt>a*)ab> will never match, since C<(?E<gt>a*)>
+For example: C<< ^(?>a*)ab >> will never match, since C<< (?>a*) >>
(anchored at the beginning of string, as above) will match I<all>
characters C<a> at the beginning of string, leaving no C<a> for
C<ab> to match. In contrast, C<a*ab> will match the same as C<a+b>,
C<a*ab> will match fewer characters than a standalone C<a*>, since
this makes the tail match.
-An effect similar to C<(?E<gt>pattern)> may be achieved by writing
+An effect similar to C<< (?>pattern) >> may be achieved by writing
C<(?=(pattern))\1>. This matches the same substring as a standalone
C<a+>, and the following C<\1> eats the matched string; it therefore
-makes a zero-length assertion into an analogue of C<(?E<gt>...)>.
+makes a zero-length assertion into an analogue of C<< (?>...) >>.
(The difference between these two constructs is that the second one
uses a capturing group, thus shifting ordinals of backreferences
in the rest of a regular expression.)
\)
}x
-which uses C<(?E<gt>...)> matches exactly when the one above does (verifying
+which uses C<< (?>...) >> matches exactly when the one above does (verifying
this yourself would be a productive exercise), but finishes in a fourth
the time when used on a similar string with 1000000 C<a>s. Be aware,
however, that this pattern currently triggers a warning message under
-B<-w> saying it C<"matches the null string many times">):
+the C<use warnings> pragma or B<-w> switch saying it
+C<"matches the null string many times">):
-On simple groups, such as the pattern C<(?E<gt> [^()]+ )>, a comparable
+On simple groups, such as the pattern C<< (?> [^()]+ ) >>, a comparable
effect may be achieved by negative look-ahead, as in C<[^()]+ (?! [^()] )>.
This was only 4 times slower on a string with 1000000 C<a>s.
in many situations where on the first sight a simple C<()*> looks like
the correct solution. Suppose we parse text with comments being delimited
by C<#> followed by some optional (horizontal) whitespace. Contrary to
-its appearence, C<#[ \t]*> I<is not> the correct subexpression to match
+its appearance, C<#[ \t]*> I<is not> the correct subexpression to match
the comment delimiter, because it may "give up" some whitespace if
the remainder of the pattern can be made to match that way. The correct
answer is either one of these:
A powerful tool for optimizing such beasts is what is known as an
"independent group",
-which does not backtrack (see L<C<(?E<gt>pattern)>>). Note also that
+which does not backtrack (see L<C<< (?>pattern) >>>). Note also that
zero-length look-ahead/look-behind assertions will not backtrack to make
the tail match, since they are in "logical" context: only
whether they match is considered relevant. For an example
where side-effects of look-ahead I<might> have influenced the
-following match, see L<C<(?E<gt>pattern)>>.
+following match, see L<C<< (?>pattern) >>>.
=head2 Version 8 Regular Expressions
Same as C<S{0,1}?>, C<S{0,BIG_NUMBER}?>, C<S{1,BIG_NUMBER}?> respectively.
-=item C<(?E<gt>S)>
+=item C<< (?>S) >>
Matches the best match for C<S> and only that.
For this grouping operator there is no need to describe the ordering, since
only whether or not C<S> can match is important.
-=item C<(?p{ EXPR })>
+=item C<(??{ EXPR })>
The ordering is the same as for the regular expression which is
the result of EXPR.
https://perl5.git.perl.org / perl5.git / blobdiff