Patient ones:

Perl surprised me again this morning. I wanted to *copy* an anonymous hash by dereferencing it, and then creating a new reference to it, leaving me with two references to two different hashes. So I did:

my $this = { 'name' => 'alex' };
my $that = \%{ $this };
# now i thought that $that would be an independent copy, but:
$that->{'name'} = 'madonna';
print $this->{'name'}; 
# oh no! i'm madonna! they both referenced the same thing

[download]

This surprised me, because I thought dereferencing would make it just a 'free-floating' anonymous hash. This works fine, but is an extra line:

my $this = { 'name' => 'alex' };
my %that = %{ $this };
my $that = \%that;
# now $that really is an independent copy, and:
$that->{'name'} = 'madonna';
print $this->{'name'}; #alex
# phew! I'm still me!

[download]

My questions are: 1) why does this happen like this? I don't really see what the difference is between the first and the second code sample. What subtlety am I missing? 2) Is there a neater way of writing the second sample, which does what I want, without using the throwaway named hash %that?

with thanks

/=\

In reply to hash dereferencing / copying by ViceRaid

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