On 3/7/06, Gavin Bowlby <[EMAIL PROTECTED]> wrote:
> So
>
> (\.)?
>
> differs from
>
> (\.?)
>
> in that there is no match in the first case, but there is a match in the
> second case?
Well, to be sure, we're talking about a successful pattern match in
either case. But the first one has a quantifi
Gavin Bowlby wrote:
So
(\.)?
differs from
(\.?)
in that there is no match in the first case, but there is a match in the
second case?
I'm still confused as to why the placement of the ? operator inside or
outside the parentheses makes a difference.
I thought the parentheses were only there
esult to a $N
variable, not to change the operations of the pattern matching itself.
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of
Tom Phoenix
Sent: Tuesday, March 07, 2006 6:51 PM
To: Gavin Bowlby
Cc: beginners@perl.org
Subject: Re: question on pattern mat
On 3/7/06, Gavin Bowlby <[EMAIL PROTECTED]> wrote:
> Could someone explain why the first case fails?
It's not failing. It's warning.
It's warning you that you're using an undefined value as if it were a
string. That's odd, Perl is saying, you seem to have expected $3 to
hold a string. But since
#!/usr/bin/perl
use warnings;
use strict;
my ($x, $y);
$x = "57 s";
# the following statements fail:
$x =~ /^(\s*)(\d+)(\.)?(\d*)(\s*)(s|S)/;
$y = $1.$2.$3;
# FAILS with "Use of uninitialized value in concatenation (.) or string"
# because $3 is undefined
# whereas these statements pass:
$x =~
