Perl modifying hash reference in subroutine - perl

Question

I am having trouble understanding the hash references and changing the hash in place, instead of returning it. I want to write a sub routine which will return a value from hash and also modify the hash. I was facing some issues while coding for it. So, I wrote the following basic code to understand modifying the hash in place.
#!/usr/local/bin/perl
#Check hash and array references
#Author: Sidartha Karna
use warnings;
use strict;
use Data::Dumper;
sub checkHashRef{
my ($hashRef, $arrVal) = #_;
my %hashDeref = %{$hashRef};
$hashDeref{'check'} = 2;
push(#{$arrVal}, 3);
print "There:" ;
print Dumper $hashRef;
print Dumper %hashDeref;
print Dumper $arrVal
}
my %hashVal = ('check', 1);
my #arrVal = (1, 2);
checkHashRef(\%hashVal, \#arrVal);
print "here\n";
print Dumper %hashVal;
print Dumper #arrVal;
The output observed is:
There:$VAR1 = {
'check' => 1
};
$VAR1 = 'check';
$VAR2 = 2;
$VAR1 = [
1,
2,
3
];
here
$VAR1 = 'check';
$VAR2 = 1;
$VAR1 = 1;
$VAR2 = 2;
$VAR3 = 3;
From the output, I inferred that, changes to hashDeref are not modifying the data in the reference. Is my understanding correct? Is there a way to modify the hash variable in place instead of returning it.

Answer 1

This is making a (shallow) copy of %hashVal:
my %hashDeref = %{$hashRef};
The hash-ref $hashRef still points to %hashVal but %hashDeref doesn't, it is just a copy. If you want to modify the passed hash-ref in-place, then work with the passed hash-ref:
sub checkHashRef{
my ($hashRef, $arrVal) = #_;
$hashRef->{'check'} = 2;
#...
That will leave your changes in %hashVal. In the array case, you never make a copy, you just dereference it in-place:
push(#{$arrVal}, 3);
and the change to $arrVal shows up in #arrVal.