And you're getting this warning because $2 is undefined.

Take a closer look at your regular expression. You only have one set of capturing parentheses on each side of the alternation, so you're only ever going to get one match. ie. $2 will always be undefined.

Cheers,
Darren

Thank you Darren. But I am still confused. $1 is matched to "this" or "that", so they will not be upper cased. $2 matched to other words other than "this" or "that" and change them to upper cases, in this example, the "and" was changed to "AND". It seems to me that the "and" was assigned to $2, why you thought that $2 will always be undefined? Thanks.

Because of the | in the middle of the regex there will never be a time where both parenthesis are used. That is, there will only be one set of parenthesis total. Ergo, $2 will never be defined.

I'm so adjective, I verb nouns!

chomp; # nom nom nom

You only have one set of capturing parentheses on each side of the alternation, so you're only ever going to get one match. ie. $2 will always be undefined.

Not true.

What is true that you get only one defined capture (if there's a match), but that one can very well be $2.

Only one of the alternates matches for any particular iteration. Because you have two captures in the regex one of them must be undefined for each match. Consider the following debugging version of your regex:

use strict;

my $chr;
$chr = "this and that";
$chr =~ s/
    (?:
        (this|that)(?{print "matched lh alt: $1 "}) # Left hand altern
+ate
        |
        (\w+)(?{print "matched rh alt: $2 "})       # Right hand alter
+nate
    )
    (?{print 'def 1' if defined ($1); print 'def 2' if defined ($2); p
+rint "\n"})
    /$1\U$2/xg;
print "$chr\n";
[download]

Prints:

matched lh alt: this def 1
matched rh alt: and def 2
matched lh alt: that def 1
this AND that
[download]

Note that only one of $1 and $2 are defined for each match.

$string = "A Perl among Camels";
$string =~ /(((\w+) (\w+)) ((\w+) (\w+)))/g;
print 
"1 is '$1'\n
2 is '$2'\n
3 is '$3'\n
4 is '$4'\n
5 is '$5'\n
6 is '$6'\n
7 is '$7'\n";
[download]

1 is 'A Perl among Camels'

2 is 'A Perl'

3 is 'A'

4 is 'Perl'

5 is 'among Camels'

6 is 'among'

7 is 'Camels'
[download]

I'm not actually sure what you're trying to do here, but I do know what's causing the error.

What it says right now is Match EITHER (this | that) OR (\w+) -> Any wordlike object including 'this' and 'that' and replace it with the first capture followed by the second capture. Do this globally (i.e. over and over until the whole string is changed).

The problem is, there's only ever one match. EITHER (this|that) matches, or (\w+) matches. Thus, $2 stays empty, thus uninitialized value.

while(<DATA>){for(split(' ',$_)){$_=reverse;print "$_ ";}};print "\b.\
+n";
__DATA__
tsuJ rehtonA lreP rekcaH
[download]

#!/usr/bin/perl --

use warnings;
use strict;

use re 'debug';

my $chr = "this and that";
$chr =~ s/(this|that)|(\w+)/$1\U$2/g; #error reported
print "$chr\n";
__END__
[download]

Consider these three examples (note that the example string is different):

>perl -wMstrict -le
"my $chr;
 $chr = \"x this and that y\";
 $chr =~ s/(this|that)|(\w+)/$1\U$2/g;
 print \"$chr\n\";
"
Use of uninitialized value in concatenation (.) or string ...
Use of uninitialized value in concatenation (.) or string ...
Use of uninitialized value in concatenation (.) or string ...
X this AND that Y
[download]

>perl -wMstrict -le
"my $chr;
 $chr = \"x this and that y\";
 $chr =~ s/(this|that)|(\w+)/\u$1\E$2/g;
 print \"$chr\n\";
"
Use of uninitialized value in concatenation (.) or string ...
Use of uninitialized value in concatenation (.) or string ...
x This and That y
[download]

>perl -wMstrict -le
"my $chr;
 $chr = \"x this and that y\";
 $chr =~ s/(this|that)|(\w+)/\u$1\E\U$2/g;
 print \"$chr\n\";
"
X This AND That Y
[download]

The number of warnings depends on which capture variable is defined and which capture variable is 'transformed' by \u or \U in the replacement string. (A replacement string without any \u or \U transformations yields five warnings.)

When both $1 and $2 are transformed in some way within the replacement string, there are no warnings. Apparently, and surprisingly (to me at any rate), the \u and \U transformations in the replacement string suppress the warnings I expected.

Offhand, I can't think why this is the case.

Update: Corrected a swapped word order.