Stupid regex problem...

[Leif K-Brooks]

I have the following Perl code. When I run it, Perl complains that $1 is
undefined. If I change (foo)|(bar) to (bar)|(foo), it works fine. I know
I could just use /(foo|bar)/, but this is just a simplified example.

use strict;
use warnings;

my $text = 'bar';
if ($text =~ /(?foo)|(bar))/) {
print $1;
}

 

[Tassilo v. Parseval]

Also sprach Leif K-Brooks:

I have the following Perl code. When I run it, Perl complains that $1 is
undefined. If I change (foo)|(bar) to (bar)|(foo), it works fine. I know
I could just use /(foo|bar)/, but this is just a simplified example.

use strict;
use warnings;

my $text = 'bar';
if ($text =~ /(?foo)|(bar))/) {
print $1;
}

You have two capturing pairs of parens of which only one will (or none
actually) can match. There is the special variable $+ (see perlvar.pod)
which holds the actual match:

my $text = 'bar';
if ($text =~ /(?foo)|(bar))/) {
print $+;
}
__END__
bar

Tassilo

 

[Andreas Kahari]

I have the following Perl code. When I run it, Perl complains that $1 is
undefined. If I change (foo)|(bar) to (bar)|(foo), it works fine. I know
I could just use /(foo|bar)/, but this is just a simplified example.

use strict;
use warnings;

my $text = 'bar';
if ($text =~ /(?foo)|(bar))/) {
print $1;
}

You never actually posed a question, so I will assume that you
meant "Why does is $1 undefined?".

Well, the regex as a whole matches, and the part that matched
the first parenthesis is stored in $1 and the part that matches
the second parentesis is stored in $2. Clearly, nothing matches
the first parenthesis, so $1 will be undefined.

 

[Leif K-Brooks]

Tassilo v. Parseval wrote:

You have two capturing pairs of parens of which only one will (or none
actually) can match.

Ok, that makes sense. I was assuming the parens would only store if they
were the ones used in the alternation. Is there any way to make that happen?