Uninitialized $1 in s///

dmitri has asked for the wisdom of the Perl Monks concerning the following question:

As a sane person, I run my Perl programs with -w. So today I stumbled across an interesting problem:

# This works fine
$a = "abc";
$a =~ s/^(?:a(\w))?(\w)$/$1:$2/; # works fine

# This produces the warning
# "Use of uninitialized value in concatenation (.)
$b = "c";
$b =~ s/^(?:a(\w))?(\w)$/$1:$2/;
[download]

I do not mind $1 to be empty; in fact, I want it to be empty if ^a\w is not there. But I only want to put the second character in $1 if the first character is "a", so

$b =~ s/^a?(\w?)(\w)$/$1:$2/; # no warning here.
[download]

won't do. The $64,000 question is, can I accomplish this in one line and not produce a warning?

Comment on Uninitialized $1 in s/// Select or Download Code

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Re: Uninitialized $1 in s/// by mirod (Canon) on Jan 02, 2002 at 09:33 UTC
This code does not generate a warning under 5.6.1, so an easy answer would be to upgrade your version of Perl. An other way is to use the `/e` option is the substitution: `$b =~ s/^(?:a(\w))?(\w)$/defined $1? "$1:$2" : ":$2"/e;` <>The `/e` option `eval`s the right part of the s/// operator, so you can test whether $1 is defined or not.	[reply] [d/l]
Re^2: Uninitialized $1 in s/// by Aristotle (Chancellor) on Jan 02, 2002 at 20:18 UTC
`s///e` is inefficient though; and this feels like a workaround more than a fix to me. Edit: on the other hand, it seems there really is no way to get around this on 5.6.0. The only thing I could come up with, though it seems slightly more efficient than `s///e`, is many times more kludgy: `$b = $1 \|\| "" . ":$2" if $b =~ /^(?:a(\w))?(\w)$/;` [download] (Which amounts to the same thing expressed in a different way..)	[reply] [d/l]
Re: Re: Uninitialized $1 in s/// by dmitri (Priest) on Jan 02, 2002 at 20:43 UTC
Aha, so it's the interpreter's fault. /e is not as cute :) I can probably do something like `$command = join(":", $command =~ /^(?:a(\w))?(\w)$/);` [download] but this requires an extra function call.	[reply] [d/l]
Re: Uninitialized $1 in s/// by blakem (Monsignor) on Jan 13, 2002 at 14:05 UTC
I've been known to rant about that warning a time or two, so I'd suggest just turning it off. Since the override is lexically scoped, this won't affect other parts of your code.... `$b = "c"; { no warnings 'uninitialized'; # turn off bogus warning for this bl +ock $b =~ s/^(?:a(\w))?(\w)$/$1:$2/; # voilla, no warning!! }` [download] -Blake	[reply] [d/l]