Perl: Passing by reference does not modify the hash - perl

Question

My understanding was that in Perl we pass hashes to functions by reference
Consider the following example, where we modify the hash in the modifyHash function
#!/usr/local/bin/perl
my %hash;
$hash{"A"} = "1";
$hash{"B"} = "2";
print (keys %hash);
print "\n";
modifyHash(\%hash);
print (keys %hash);
print "\n";
sub modifyHash {
my $hashRef = #_[0];
my %myHash = %$hashRef;
$myHash{"C"} = "3";
print (keys %myHash);
print "\n";
}
The output of this script is:
AB
ABC
AB
I would have expected it to be:
AB
ABC
ABC
...as we pass the hash by reference.
What concept am I missing here about passing hashes to functions?

Answer 1

That's because when you do my %myHash = %$hashRef;, you're taking a copy of the dereferenced $hashref and putting it into %myHash which is the same thing as my %myHash = %hash;, so you're not working on the referenced hash at all.
To work on the hash specified by the reference, try this...
sub modifyHash {
my $hashRef = $_[0];
$hashRef->{"C"} = "3";
print (keys %$hashRef);
print "\n";
}
As pointed out by ThisSuitIsBlackNot in the comments below, #_[0] is better written as $_[0]. You should always be using use strict; and use warnings;, as this would have been caught. Because you're sending in a reference, you could also have used my $hashRef = shift;.

Answer 2

The problem is with the assignment:
my %myHash = %$hashRef;
This is akin to saying:
$x = 5;
$y = $x;
You're not setting $y to reference the same spot in memory, you're just giving the value of $x to $y. In your example, you're creating a new hash (%myHash) and giving it the value of the hash stored at $hashRef. Any future changes are to the new hash, not the original.
If you want to manipulate the original, you should do something like:
${$hashRef}{"C"} = "3";
or
$hashRef->{"D"} = 4;
There might be a more elegant way of doing it, but as far as I know you want to work with the hash reference.