I was recently helping someone debug code in which the pass a hash to a subroutine by reference. They had done something like:

%foo = ( a => 1, b => 2, c => 3 };

sub mysub {
  my %x = shift;
...
}
[download]

It was clear to me that the assignment a scalar reference directly to a hash variable would result in a hash containing a single pair, with key the actual reference value, and value undef. I usually shift onto a scalar variable and then use a dereference operator (->) in the sub when I pass hashes around. The person I was helping wanted to keep the hash variable %x, however. So I tried a couple things and eventually got this to work:

my %foo = ( a => 1, b => 2, c => 3 );

mysub(\%foo);

sub mysub {
  my %x = %{scalar shift};

  foreach ( sort keys %x ) {
    print "$_ -> $x{$_}\n";
  }
}
[download]

That %x = %{scalar shift} is a pretty nifty little incantation. I tried to figure out why it works. Based on my experience, clearly this works:

my %foo = ( a => 1, b => 2, c => 3 );

mysub(\%foo);

sub mysub {
  my $x = shift;
  my %x = %{$x};

  foreach ( sort keys %x ) {
    print "$_ -> $x{$_}\n";
  }
}
[download]

However, when I tried to simplify the code above, I was started that this does NOT work:

my %foo = ( a => 1, b => 2, c => 3 );

mysub(\%foo);

sub mysub {
  my %x = %{shift};

  foreach ( sort keys %x ) {
    print "$_ -> $x{$_}\n";
  }
}
[download]

The above code prints out nothing. What is magic about adding that "scalar"???

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