Peter Daum am Samstag, 5. August 2006 18:45: > Hi, Hallo Peter
> when trying to process continuation lines in a file, I ran > into a weird phenomenon that I can't make any sense of: > > $s contains a line read from a file, that ends with a backslash > (+ the newline character), so > > $s='abc \ > '; > > $s =~ /^(.*)$/; print $1; # prints "abc \" as expected > > > If the line ends with a backslash, I don't want to include it > in the grouping, thus I use: > > $s =~ /^(.*[^\\])(\\)?$/; print "1: '$1', 2: '$2'"; You can see what matches what by adding additional catching () : use warnings; my $s =~ /^ (?: (.*) ([^\\]) ) (\\)? $ /x; print "1: <$1>, 2: <$2>, 3: <$3>"; This shows: 1: <abc \>, 2: < >, 3: <> ^ ¦ this is not a citation '>' The greedy .* matches as much as possible on the line: including '\' The [^\\] matches the last "char" that is not a literal '\': that's the newline. The (\\)? is optional and matches nothing, since nothing is left on the line. > I would expect $1 to hold "abc " and $2=="\\", but instead, > the first grouping holds everything including the backslash > and the following newline, while $2 is left undefined. > > The perlre manpage says: > > . Match any character (except newline) > > $ Match the end of the line (or before newline at the end) > > but in my case the "." obviously matched the newline at the end. > > When I do a chomp($s) first, everything behaves as expected, > while a "/m" at the end of the regular expression doesn't > make any difference. > > Does anybody have an explanation what is going on here? > > Regards, > Peter Daum Hilft Dir das weiter? Dani -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] <http://learn.perl.org/> <http://learn.perl.org/first-response>
