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

Dear monks, After being away from Perl for a year, i'm back, and i seek new wisdom :)

In the .Net platform an object is stored in the heap. When passing the object 'as value' and not as a reference, the object itself isn't copied, you still get a reference to it (instead of a fresh copy).

If you want a copy (like items on the stack), you need to deep copy (clone) the object by hand.

I wonder how this is done in Perl, since i could not find any info on it.

Thanks again :)

edit:
as just stated by jkva (and the others below (thanks) ).... an object in Perl already IS a reference... So it stays a reference as it's passed :).
  • Comment on VB-ish behaviour byVar or byRef of objects in Perl

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Re: VB-ish behaviour byVar or byRef of objects in Perl
by moritz (Cardinal) on Dec 11, 2007 at 14:54 UTC
    Since an "object" in Perl 5 is a blessed reference, you get the same behaviour.

    You can deep clone an object with Storable::dclone.

[reply]
Re: VB-ish behaviour byVar or byRef of objects in Perl
by holli (Abbot) on Dec 11, 2007 at 14:47 UTC
    Clone


    holli, /regexed monk/
[reply]
[d/l]
Re: VB-ish behaviour byVar or byRef of objects in Perl
by locked_user sundialsvc4 (Abbot) on Dec 11, 2007 at 15:37 UTC

    This behavior is quite common and it's implemented and used quite heavily in (current versions of...) Perl. See perldoc perlref et al.

    One of the nice concepts that's used a lot in these languages is this one:   the thing that is passed “is a” reference. In other words, it no longer matters how you pass it... rather, the thing that is passed is a magical thing that will give access to whatever it is “referencing.” It's a subtle distinction but very useful.

    Like the other languages, Perl automagically provides you with reference-counted objects that are garbage collected for you.

Re: VB-ish behaviour byVar or byRef of objects in Perl
by ikegami (Patriarch) on Dec 11, 2007 at 18:56 UTC

    The same applies to non-objects as well. Everything is passed by reference in Perl. 

    sub func { $_[0] = "bar"; }

my $s = "foo";
func($s);
print("$s\n");  # bar

    [download]
[reply]
[d/l]
      Everything is passed by reference in Perl.

      I know it's true, but the common ways to handle subroutine arguments sort of make it untrue.

      sub fiddle {
    my ($arg) = @_;
    $arg = "waffle";
}

sub faddle {
    my $arg = shift;
    $arg = "waffle";
}

sub muddle {
    $_[0] = "waffle";
}

my $value = "pancake";
print "$value\n"; # Prints "pancake"
fiddle($value);
print "$value\n"; # Prints "pancake"
faddle($value);
print "$value\n"; # Prints "pancake"
muddle($value);
print "$value\n"; # Prints "waffle"

      [download]

      Shifting and slicing @_ into subroutine lexicals will make copies (which only matters for simple non-reference values). I know, I know. This is obvious stuff I'm talking about. But when somebody hears that everything is passed by reference in Perl, they also need to hear that it's a behavior that often gets bypassed in idiomatic code.

[reply]
[d/l]

        Consider that it's because of the passed-by-aliasing behavior that those are the prevailing idoms . . . :)

        The cake is a lie.
        The cake is a lie.
        The cake is a lie.

[reply]

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