Clint, thanks for answering. I am actually going to do exactly what you said, however, I am trying to find where it points to a value so I can reinit it into a hash. I ended up doing it in strings, and its working fine now, I take it step by step and if at any point it points to a value it simply gets replaced by a hash as needed. I am little stuck on the following code perhaps you could help me out.
sub addTo()
{
my $location = $_[0];
my $data = $_[1];
my @keys = split(/\//, $location);
my $key;
my $current = $database;
while($key = shift(@keys))
{
$current->{$key} = {} unless (exists $current->{$key} && ref($curr
+ent->{$key}) eq "HASH");
$current = $current->{$key};
}
#I AM STUCK HERE -----
$current = $data;
}
Since $current is a pointer, I understand that by setting it to $data it doesn't change the value of what its referencing but changes $current to equal the scalar value $data.
I tried the following snippet
my $pointer = \$database->{"2002"}->{"Gustav"}->{"General"}->{"Overvie
+w"};
${$pointer} = "nasdfaull";
This worked well, and actually changed the value it was referencing. However, when I tried to do the same referencing technique below with $current = $database, I would get stuck on the unless line, telling me that I didn't have a hash reference. The addTo method definitely works if I keep it as it now but switch the
$current = $data code with
$current->{$key} = $data;
I would really like to make it work without having to switch that though.
-Constantin | [reply]
[d/l]
[select] |
There is a bit of nomenclature here that is not working in your favor. Perl does not have a concept of "pointers", although it does use "references" a lot (there is a not so subtle distinction - in Perl objects are reference counted). Perl allows references to pretty much anything, including references so you can have what may be considered indirect references, but they still ain't pointers.
I suspect a good dose of reading is what you most need at this point. Take a look at perldata and perllol. You may find the Tutorials section Data Types and Variables helps a lot too.
Assignment always replaces whatever was previously contained by a variable. It doesn't matter what "type" of thing the variable previously contained, it gets replaced.
DWIM is Perl's answer to Gödel
| [reply] |
I ended up doing $current->{$key} = $data, and after reading your post I think I see how I confused myself with the pointer/references issue, Perl not having pointer, only references. Thanks for the help guys.
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