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

Evening! I'm trying to append to a scalar using a reference. I'm doing the same thing for an array which works, but this doesn't - isntead, it seems to have passed the reference name to my variable: 
sub file_expansion {
        my ($output, $input) = @_; 
        my $test = $output;     # For testing whether I need to append
+ to an array or a scalar#
        while (<@$input>) { 
                if (-e $_) {
                        open (FILE, "< $_");
                        my @lines = <FILE>;
                        foreach my $line (@lines) {
                                $_ = $test;
                                if (/SCALAR/) {
                                        $output .= $line;
                                        
                                }
                                else {
                                        push @$output, $line;
                                }
                        }
                }
                else {
                        print "<<ERROR: File \'$_\' doesn't exist or c
+an't be read.>>";
                }
        }
}

@temp_fyle = "/home/monkey/list_of_commands"; (looks like below:
interface blah 
ip address 1.2.3.4 255.255.255.255
)

&file_expansion(\$template,\@temp_fyle);

[download]
Ok, what I'd expect is that printing $output within the sub will result in the contents of the file being in there. What I'm finding, however, is that I get something like "SCALAR(0x828c810)interface blah ip address 1.2.3.4 255.255.255.255" And after exiting the loop, $template is empty. Obviously what's happening is that I've not referenced it properly and the $output variable actually has the scalar reference as its value rather than being a reference to the original scalar. What have I done wrong?
toeslikefingers.com - because you've got too much time on your hands

Replies are listed 'Best First'.
Re: referencing question
by gjb (Vicar) on Nov 06, 2003 at 12:59 UTC
    $$output .= $line;
    should do the trick, you forgot to dereference the reference to the scalar. For the list reference, you correctly used @$output though.

    Hope this helps, -gjb-

[reply]
[d/l]
[select]
      Yup, a guy across the room just explained it. Finally getting to grips with all this referencing business. Also figured out that I should be using ref() rather than my dodgy 'assign reference to a string and test that' business with the '$test' variable.
      toeslikefingers.com - because you've got too much time on your hands
[reply]
Re: referencing question
by hmerrill (Friar) on Nov 06, 2003 at 17:05 UTC
    When passing parameters, you want to pass scalars - which means you want to pass: 
       * scalars as themselves
   * each array as an array reference (a reference is a scalar)
   * each hash as a hash reference (a reference is a scalar)

    [download]
    so 
      1. there is no reason to pass a reference to
     $template(a scalar) to the subroutine - why not just
     pass $template?  Forget about the reference -
     then in the subroutine you can do

        my ($output, $input) = @_;

     and just use $output in the subroutine, instead
     of having to use $$output.

  2. when you pass \@temp_fyle in the call to
     subroutine file_expansion, that's fine.  This is
     just my preference, but IMHO it helps keep your
     code simpler and more understandable if you name
     variables consistent with what they contain.  For
     example, in subroutine file_expansion, name the
     imcoming parameter that is a reference to the array
     what it is - a reference, like this:

         my ($output, $input_arrayref) = @_;

     or maybe even better

         my ($output, $temp_fyle_arrayref) = @_;

     Then, since you can actually use the reference to point
     at array elements (instead of having to dereference
     the array reference into an array), you can do something
     like this

         print "first element is $temp_fyle_arrayref->[0]\n";

     to refer to the 1st element of the array, instead of
     having to do
 
         my @new_array = @$temp_fyle_arrayref;
         my $first = @new_array[0];
         print "first element is $first\n";

     which takes more memory since you are creating another
     array(@new_array).

    [download]
    HTH.
[reply]
[d/l]
[select]
      hmerrill: I was under the impression that if you had very large scalar (e.g. $template contains a 100kb template) it was more efficient to pass a reference than to pass the whole scalar, since no memory is being copied. Or am I confused on this point?
[reply]
        No, your impression is probably correct - I've never really used scalars with really *large* values, so that's probably why I didn't think of that. In that case, you *should* probably pass a reference to the scalar like you were doing in the first place. Sorry if I confused you.
[reply]

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