can scalars be passed by reffernence

Sure can.

resulting regular expression in the library untaint them?

You need to dereferance the value and untaint it:

sub untaint_ref 
{
    my $in = shift;

    $$in =~ /(.*)/;

    return $1;
}

my $val = 12345;
$val = untaint_ref(\$val);
[download]

Obviously, you want a better untainting method than /(.*)/.

----
I wanted to explore how Perl's closures can be manipulated, and ended up creating an object system by accident.
-- Schemer

Note: All code is untested, unless otherwise stated

at this point it seems clear to me that i have no idea what i am doing.... after messing with this for the last hour the best i have gotten is

#!/usr/bin/perl -w
use strict;
my $thingy = "No Change\n";
print $thingy;

&change($thingy);

print $thingy;

sub change
{
  my $in = shift;
  $$in = "Changed\n";
}
[download]

I get the following error

Can't use string ("No Change") as a SCALAR ref while "strict refs" in 
+use at asdf.pl line 12
[download]

It seems like it is doing some sort of weird passing by value thing to me... but then I have no idea what I am doing (haven't you noticed...sigh) I read through perlref and it mentioned this problem existed but I could discern no answer also the 2 perl books I have on hand made similar statments with no answer I could find.

and I forgot to login, blast I guess today is not my day.......

It seems like it is doing some sort of weird passing by value thing to me

Yes, but it's not weird, it's doing exactly what you're telling it -- passing a string that you're then trying to dereference. Look at hardburn's post and you'll see the critical difference is the backslash preceding the scalar.

&change($thingy); # passes by value
&change(\$thingy); # creates and passes a *reference*
[download]

---

"The dead do not recognize context" -- Kai, Lexx

First, you should avoid the ampersand syntax unless you have a specific reason to use it.

Second, you're passing by value, not by referance. When you get to the $$in = "Changed\n"; part, perl tries to use $in as a symbolic ref. Symbolic refs are (rightfully) illegal with strict 'refs' turned on, hence the error message. To pass by a hard referance, you need to change how you call the sub to change(\$thingy);. Notice the '\' before the '$', which is saying "make a referance to $thingy". No change should be necessary to the subroutine itself.

----
I wanted to explore how Perl's closures can be manipulated, and ended up creating an object system by accident.
-- Schemer

Note: All code is untested, unless otherwise stated

1. can scalars be passed by reference?
   yes. any data type can be passed by reference, be it scalar, array, hash, glob, etc. though, usually scalars aren't passed as a reference, because they're so easy to pass around as they are.
2. would the resulting regular expression in the library untaint them?
   if you mean the 'corresponding' regular expression, then yes. but you must first dereference the reference. for more on that, take a look at perlref.

for more tainting info, there's a good tutorial that includes tainting at Ovid's Web Programming with Perl course

~Particle *accelerates*