why can't I collapse reference variables?
Mark_Galeck
days there is no response, so maybe it is not a beginners question?
Nah, it must be :)
--------------------------------
If I can do this:
$ref = \@foobar;
print @$ref;
then why can't I do this:
print @\@foobar;
------------------------------------
I have this kind of problem often, let me give another example, I
think it is similar, let me know if not.
Why are all the printouts different:
@foobar = ();
print \@foobar; #prints as reference to ARRAY
$foobar = \();
print $foobar; # prints as reference to SCALAR
print \(); #prints as array of references to SCALAR, in this case
empty array
James E Keenan
On Nov 27, 2009, at 10:48 PM, Mark_Galeck wrote:
> $ref = \@foobar;
> print @$ref;
>
> then why can't I do this:
>
> print @\@foobar;
use strict;
use warnings;
> ------------------------------------
> I have this kind of problem often,
If you're having this problem often, it's probably because you're not
using strictures and warnings often.
G. Wade Johnson
On Fri, 27 Nov 2009 19:48:09 -0800 (PST)
Mark_Galeck <mark_galeck...@yahoo.com> wrote:
> Hello, the following question I posted on perl.beginners but for a few
> days there is no response, so maybe it is not a beginners question?
> Nah, it must be :)
This question bounces on some subtleties in Perl's references and
syntax.
> --------------------------------
> If I can do this:
>
> $ref = \@foobar;
> print @$ref;
Yes.
> then why can't I do this:
>
> print @\@foobar;
This appears to be a syntax error, probably having to do with the @\@
not parsing as you expect. If you re-write it as
print @{\@foobar};
is works as you probably intend. Unfortunately, I cannot imagine why
you would want to do that.
> ------------------------------------
> I have this kind of problem often, let me give another example, I
> think it is similar, let me know if not.
This is a completely different issue.
> Why are all the printouts different:
>
> @foobar = ();
> print \@foobar; #prints as reference to ARRAY
As it should.
> $foobar = \();
> print $foobar; # prints as reference to SCALAR
Although many people misunderstand it, () does not create a list. The
parens only provide grouping. In this case, you are taking a reference
to a grouped nothing and perl is interpreting that as a request for
a reference to an undef scalar.
> print \(); #prints as array of references to SCALAR, in this case
> empty array
Correct.
I would definitely suggest spending quality time with perlref and
perlreftut. They will help you work through the syntactic oddities in
references. There is a really good discussion of some of this at the
Perl Monastery (http://perlmonks.org/?node_id=779217)
Many people develop habits that reduce some of the confusion. I only
derefence references in one of two ways (out of several).
* If I need the whole item, I use the sigil and curlies. I never use
just the derefencing sigil.
@{$foo} not @$foo
It reads better to me and is less prone to the kinds of syntax attack
you had above.
* If I need an element of a referenced array or hash, I use the arrow
notation.
$foo->{'a'} not ${$foo}{'a'}
I find that the arrow helps me later in reading the code. The visual
reminder of the reference simplifies my own code reading.
I would definitely suggest you check out the Perl Monastery. It is
a great community for helping with questions like this.
G. Wade
--
Against logic there is no armor like ignorance. -- Laurence J. Peter
Vlado Keselj
On Fri, 27 Nov 2009, Mark_Galeck wrote:
> Hello, the following question I posted on perl.beginners but for a few
> days there is no response, so maybe it is not a beginners question?
> Nah, it must be :)
I cannot provide brief explanations, but this may help:
> --------------------------------
> If I can do this:
>
> $ref = \@foobar;
> print @$ref;
>
> then why can't I do this:
>
> print @\@foobar;
use: print @{ \@foobar };
It seems to work.
> ------------------------------------
> I have this kind of problem often, let me give another example, I
> think it is similar, let me know if not.
>
> Why are all the printouts different:
>
> @foobar = ();
> print \@foobar; #prints as reference to ARRAY
This is OK.
>
> $foobar = \();
> print $foobar; # prints as reference to SCALAR
use: $foobar = []; print $foobar;
and you get the output as before.
> print \(); #prints as array of references to SCALAR, in this case
> empty array
use: print [];
For creating a reference to a literal array, use: [1, 2, 3] instead of
\(1,2,3)
Cheers,
--Vlado