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Think about Loose Coupling
 
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This does not affect the real %hash. By de-referencing it like that, you're essentially creating a copy. The item on the right-hand-side is now a real hash (by virtue of wrapping it in %{...}), not a reference. Thus, you are effectively doing this:
%hash1 = %hash2;
This does not create an alias, but a copy. Changes to %hash1 have no effect on %hash2.

And no, you don't have to put parens around your split call. Perl does the correct thing here, but it may be more readable to others if you write it with parens anyway. Here is some test-case code you can try yourself:

sub test { my $hash1 = shift; my %hash2 = %{shift}; $hash1->{three} = 3; $hash2{three} = 3; } my %a = (one => 1, two => 2); my %b = (one => 1, two => 2); &test(\%a, \%b); print join(", ", keys %a), "\n"; print join(", ", keys %b), "\n"; # three, two, one # two, one

In reply to RE: Answer: how do I use a hash of hashes reference to a function ? by Fastolfe
in thread how do I use a hash of hashes reference to a function ? by Anonymous Monk

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