G'day harangzsolt33,

"I would like to point out a distinction."

You have failed to do that. The 108 lines of code that you've posted contain no references to functions, subroutines, or any other code.

Calling a variable $CODE_REF does not make it a coderef. $CODE_REF is not any kind of reference; it's just a string:

$ perl -MO=Deparse -e 'my $CODE_REF = \"if (\$c == 64) { \$MODE = prin
+t \"\\n\"; }" . " else { print chr(\$MODE = \$MODE & 1 ? 32 : 35) x \
+$c; } ";'
my $CODE_REF = 'SCALAR(0xa00038be0) else { print chr($MODE = $MODE & 1
+ ? 32 : 35) x $c; } ';
-e syntax OK
[download]

You can explicitly check it yourself:

$ perl -e 'my $CODE_REF = \"if (\$c == 64) { \$MODE = print \"\\n\"; }
+" . " else { print chr(\$MODE = \$MODE & 1 ? 32 : 35) x \$c; } "; my 
+$ref = ref $code_ref; print $ref ? $ref : "not a ref";'
not a ref
[download]

Or let Perl do it for you:

$ perl -Mstrict -e 'my $CODE_REF = \"if (\$c == 64) { \$MODE = print \
+"\\n\"; }" . " else { print chr(\$MODE = \$MODE & 1 ? 32 : 35) x \$c;
+ } "; $CODE_REF->();'
Can't use string ("SCALAR(0xa00038c70) else { print"...) as a subrouti
+ne ref while "strict refs" in use at -e line 1.
[download]

"Is there any other way to do it?"

You can modify a caller's variables by passing references to those variables. Here's a very simple example:

#!/usr/bin/env perl

use strict;
use warnings;

my ($x, $y) = qw{X Y};

print "BEFORE:\n";
print "\$x[$x] \$y[$y]\n";
mod_callers_vars(\$x, \$y);
print "AFTER:\n";
print "\$x[$x] \$y[$y]\n";

sub mod_callers_vars {
    my ($x_ref, $y_ref) = @_;
    $$x_ref = 'A';
    $$y_ref = 'B';
    return;
}
[download]

Output:

BEFORE:
$x[X] $y[Y]
AFTER:
$x[A] $y[B]
[download]

I see LanX has shown you how to use closures (which could be appropriate for more complex scenarios).

In "Re^4: Aren't there code refs as well as function refs?", you wrote:

"... you decide to call each function with 30 or so arguments."

That would be a very poor decision. Instead, use a single hashref argument.

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