I currently have the following
# $dog, $cat, $rat are all hash refs
my %rethash = ('success' => 'Your Cool');
my %ref ={ 'dog' => $dog, 'cat' => $cat, 'mouse' => $rat,
'chicken' => '' };
my $perlobj = ( \%ref,\%rethash );
When $perlobj is dumped this is the result
$VAR1 = {
'success' => 'Your Cool'
};
However when warnings are enabled I get the following message
Useless use of reference constructor in void context at ..
I realize there is something terribly wrong with how %ref is assigned using {}, What is wrong with this code? I can't seem to get rid of this warning....
EDIT: Ok I think I figured out whats going on,
my $perlobj = ( \%ref,\%rethash );
This does not merge but results in $perlobj becoming a reference to %rethash, this is obvious after reading your responses.
What RobEarl is saying is correct. I'll give an explanation of that and add some more stuff.
Your variable name %ref and the fact that you are using {} kinda implies you want a reference here.
Let's take a look what value we will have in %ref. Consider this example.
use strict; use warnings;
use Data::Printer;
my %foo = { key => 'value' };
p %foo;
This will throw a warning Reference found where even-sized list expected on my Perl 5.20.2. The output will be:
{
HASH(0x7e33c0) undef
}
It's a hash with a hashref as the key and undef as a value. HASH(0x07e33c0) is what you get when you look at a hash reference without dereferencing it. (The {} are there because Data::Printer converts the hash to a hashref).
Back to your code, the correct sigil for a reference is $. It does not matter what kind of reference it is. The reference is always a scalar (a pointer to the place in memory where the hash/array/something) is stored.
my $ref = {
dog => $dog,
cat => $cat,
mouse => $rat,
chicken => '', # maybe this should be undef?
};
Now you've got a hashref with the values of $dog, $cat, $rat and an empty string.
Now you're assigning a variable named $perlobj, which implies it's an object. Instead you are assigning a scalar variable (the $ makes it a scalar) with a list. If you do that, Perl will only assign the right-most value to the variable.
my $foo = (1, 2, 3); # $foo will be 3 and there's a warning
You are assigning a list of two references. The first one is disregarded and only \$rethash gets assigned. That works because conveniently, $perlobj is a scalar, and references are also scalars. So now $perlobj is a reference of %rethash. That's why your Data::Dumper output looks like %rethash.
I'm not sure what you want to do, so I cannot really help you with that. I suggest you read up on some stuff.