When you call unfold_hash($data, $result_hr), the special variable @_ contains two elements, each that are aliases to the things you passed in. The first element is an alias to the variable that holds a reference to a datastructure. The second element is an alias to a variable that is not yet defined.

Aliasing no longer exists in these new variables. Yet you can modify the datastructure referred to by the reference that $raw_hr, and that change propagates outward, because they refer to the same data structure. On the other hand, if you assign a new reference to $raw_hr, it would not have anything to do with the $result_hr variable because no aliasing is in effect, and $result_hr does not know of this reference.

When you autovivify a hashref and assign it into $raw_hr there is no mechanism by which that hashref would propagate outward; there is no aliasing in effect here, and the reference was unknown outside of the subroutine. In the case where you set $result_hr = {}, you create a reference to an empty data structure, and the subroutine obtains a copy of that reference (not of the structure), which it can then manipulate such that changes to the structure propagate outward. ...this is high risk behavior, however, as it changes the data structure... anyone referring to it will have the same change, because it is the same structure.

Fantastic explanation! The best documentation I have been able to find about this is perlsub, but I would love to see any more details if you have them.

Here is the OP's code working as he expected:

perl -e '
> use strict;
> use warnings;
> use Data::Dumper;
>
> my $data = {status=>"ok","message-type"=>"member","message-version"=
+>"1.0.0"};
> my $result_hr;
>
> unfold_hash($data, $result_hr);
> print Dumper $result_hr;
>
> sub unfold_hash {
>     for my $k ( keys %{ $_[0] } ) {
>             $_[1]{$k} = $_[0]{$k};
>     }
>      print "unfold_hash ", $_[1] ? scalar %{ $_[1] } : 0, "\n";
> }
> '
unfold_hash 3
$VAR1 = {
          'message-version' => '1.0.0',
          'message-type' => 'member',
          'status' => 'ok'
        };
[download]

πάντων χρημάτων μέτρον έστιν άνθρωπος.

Simplifying a bit, references point to a memory location of a data structure. In your first piece of code, my $result_hr; will be undef because it isn't initialized. In your sub, you copy the elements of @_ into $raw_hr and $res_hr, so now $res_hr is just a copy of the original undef. In other words, it doesn't point anywhere! And $res_hr has no way to affect $result_hr. There is a hash being autovivified, but it is restricted to the scope of the sub.

When you write my $result_hr={};, now what is being passed into the sub is a reference that points to an anonymous hash. The reference still gets copied into $res_hr, but the result of that is just two references pointing at the same data structure. The hash isn't being autovivified in sub, instead the sub is modifying the anonymous hash pointed to by both $result_hr and $res_hr. When the sub returns, your main code still has access to the anonymous hash via the reference stored in $result_hr.

Note that there is a way to modify $result_hr from within the sub: use $_[1] directly, instead of $res_hr, because the elements of @_ are aliases to the original arguments. However, I wouldn't normally recommend this way of modifying arguments, because that they are in effect return values can be confusing to users of the API. It's more common to handle this via regular return values.

(Update before posting: I see davido has explained the same thing, with different wording - TIMTOWTDI :-) )

There is indeed more than one way to explain references, and to date I don't believe anyone has found an optimal way. ;)




Dave

There is indeed more than one way to explain references, and to date I don't believe anyone has found an optimal way. ;)

Yep, in my experience it differs from person to person which kind of an explanation will make it "click" :-)

(And just to be clear, your explanation is very good and my post was in no way intended to be a commentary on it - hence the TIMTOWTDI comment!)

#!/usr/bin/perl

# https://perlmonks.org/?node_id=1224994

use strict;
use warnings;
use Data::Dumper;

my $data = {status=>"ok",'message-type'=>"member",'message-version'=>"
+1.0.0"};
my $result_hr;

unfold_hash($data, $result_hr);
print Dumper $result_hr;

sub unfold_hash {
    my ( $raw_hr ) = @_;
    for my $k ( keys %$raw_hr ) {
            $_[1]->{$k} = $raw_hr->{$k};
    }
     print "unfold_hash ", $_[1] ? scalar %{$_[1]} : 0, "\n";
}
[download]

Outputs:

unfold_hash 3
$VAR1 = {
          'message-version' => '1.0.0',
          'status' => 'ok',
          'message-type' => 'member'
        };
[download]

Good question. The others explained it well already but here is another hint. (You weren't actually passing a reference is why it didn't work.)

use strict;
use warnings;

my $data = {status=>"ok",'message-type'=>"member",'message-version'=>"
+1.0.0"};

my $result_hr;
ref($result_hr) ? print "ref($result_hr)\n" : print "\$result_hr is no
+t a reference.\n" ;
$result_hr = {};
ref($result_hr) ? print "ref($result_hr)\n" : print "\$result_hr is no
+t a reference.\n" ;

__DATA__
$result_hr is not a reference.
ref(HASH(0x3ddb70))
[download]

sub unfold_hash {
    my ( $raw_hr ) = @_;
    my %res;

    for my $k ( keys %$raw_hr ) {
        $res{$k} = $raw_hr->{$k}; 
    }

    print "unfold_hash ", %res ? scalar %res : 0, "\n";
    \%res;
}

$result_hr= unfold_hash($data);
print Dumper $result_hr;
[download]

Is the question for the sake of understanding reference and sub-routine? If not, the code is basically just merging one hash to another.

 my $result_hr = { %$data };
[download]