Not a code reference in Perl class - perl

Question

I'm stumped. I'm new to Perl and after reading some articles, I still can't figure this one out. It's a very small class.
package Haha;
sub new {
$class = shift;
$self = {
path => shift
};
bless $self, $class;
return $self;
}
sub setPath {
my ($self, $new_path) = shift;
$self->(path) = $new_path if defined $new_path;
return $self->(path);
}
sub getPath {
my $self = shift;
return $self->(path);
}
1;
And I used it like this:
use lib 'lib';
use Haha;
my $new_excel = new Haha("sample path");
print $new_excel->getPath() ;
<>;
Class Haha line 23 raises the "Not a code reference" error.
The line that says return $self->(path);

Answer 1

Your class (like most Perl classes) is implemented on top of hashes. When you create a new object in your constructor, you do it like this:
sub new {
$class = shift;
$self = {
path => shift
};
bless $self, $class;
return $self;
}
The line $self = { ... } creates an anonymous hash and stores a reference to that hash in $self. So, $self is a hash reference. Which means that you should access its contents using hash syntax. So your accessor and mutator methods are wrong.
sub setPath {
my ($self, $new_path) = shift;
$self->(path) = $new_path if defined $new_path;
return $self->(path);
}
You are using parentheses, not braces, to access the path value in your hash. The line:
$self->(path) = $new_path if defined $new_path;
Should be:
# Note: braces, not parentheses
$self->{path} = $new_path if defined $new_path;
And the line:
return $self->(path);
Should be:
# Note: braces, not parentheses
return $self->{path};
You need to make a similar fix to getPath().
Unfortunately, the syntax $reference->($value) is completely valid. It means "call the subroutine that you have a reference to in $reference, passing it $value". But, of course, this requires $reference to contain a subroutine reference, not a hash reference.
A few other suggestions.
Always use strict and use warnings.
Indirect object notation ($new_excel = new Haha("sample path")) is likely to burn you at some point. Please use $new_excel = Haha->new("sample path") instead.
Your line my ($self, $new_path) = shift doesn't do what you think it does. You want my ($self, $new_path) = #_.

Answer 2

path is an attribute of the object, use curly brackets:
sub getPath {
my $self = shift;
return $self->{path};
}
In the sub setPath, the variable $new_path is never assigned, use instead:
sub setPath {
my ($self, $new_path) = #_;
$self->{path} = $new_path if defined $new_path;
return $self->{path};
}