Front page | perl.perl5.porters |
Postings from February 2000
On Sun, 27 Feb 2000 14:42:27 MST, Tom Christiansen wrote:
>See the problem? Here's a patch for File::Basename, but there are
>many more to go. Anywhere that a pattern is applied to a filename,
>the pattern must be closely inspected, and the following measures
>taken:
>
> If you find a "." to mean "any character", the pattern must end
> in /s.
>
> If you find a "^" to mean "start of string", the pattern must
> also end in /s, or the "^" altered to a "\A".
>
> If you find a "$" to mean "end of string", the "$" altered to
> a "\z". Not "\Z", but "\z".
But \z doesn't appear to work properly. :-(
% ./perl -we 'print "[$1]\n" if "foo\n" =~ /(.+)\z/'
% ./perl -we 'print "[$1]\n" if "foo\n" =~ /(.+)\z/s'
[foo
]
% ./perl -we 'print "[$1]\n" if "foo\n" =~ /(.+)$/s'
[foo
]
%
So \z only appears to work when /s is specified, but $ works the
same with /s. What good is \z then, I wonder?
I guess $ would be different from \z when there's something after $
that can match the newline, but then, even that doesn't seem to work.
% ./perl -we 'print "[$1]\n" if "foo\n" =~ /(.*)$\n/'
%
Hmm. Oh, that appears to be parsed as C<"(.*)" . ${\} . "n">. Putting
a space between the $ and the \ makes it work.
% ./perl -we 'print "[$1]\n" if "foo\n" =~ /(.*)$ \n/x'
[foo]
Another issue with \z: /foo\z\n/ should never match, but why isn't it
a compile-time error?
>In fact, it might be a good idea to just put /s on everything, but
>you still have the \z issue.
I'll say.
>This is *seriously* bad. Not only did the perl version miss the
>stuff hidden in "..\n" directory, it proceeded to run back up the
>tree!
>
>Here's File::Basename. I've got to back to some other stuff now.
>There's a lot to be fixed. The standard Perl libraries need to all
>be subjected to a complete code inspection for these issues.
I don't think we can use \z until it actually works.
Sarathy
gsar@ActiveState.com
Thread Previous
|
Thread Next