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From: PerlFAQ Server on 23 Apr 2010 00:00
This is an excerpt from the latest version perlfaq4.pod, which
comes with the standard Perl distribution. These postings aim to
reduce the number of repeated questions as well as allow the community
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perlfaq is at http://faq.perl.org .

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4.68: Why does passing a subroutine an undefined element in a hash create it?

(contributed by brian d foy)

Are you using a really old version of Perl?

Normally, accessing a hash key's value for a nonexistent key will *not*
create the key.

my %hash = ();
my $value = $hash{ 'foo' };
print "This won't print\n" if exists $hash{ 'foo' };

Passing $hash{ 'foo' } to a subroutine used to be a special case,
though. Since you could assign directly to $_[0], Perl had to be ready
to make that assignment so it created the hash key ahead of time:

my_sub( $hash{ 'foo' } );
print "This will print before 5.004\n" if exists $hash{ 'foo' };

sub my_sub {
# $_[0] = 'bar'; # create hash key in case you do this
1;
}

Since Perl 5.004, however, this situation is a special case and Perl
creates the hash key only when you make the assignment:

my_sub( $hash{ 'foo' } );
print "This will print, even after 5.004\n" if exists $hash{ 'foo' };

sub my_sub {
$_[0] = 'bar';
}

However, if you want the old behavior (and think carefully about that
because it's a weird side effect), you can pass a hash slice instead.
Perl 5.004 didn't make this a special case:

my_sub( @hash{ qw/foo/ } );



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