Hashes must be passed by reference, which means they're nothing more than a single scalar value (in terms of an arguement to a subroutine) If you don't pass its reference, instead just do foo(%hash) then what you're really passing in is an even paired list. HTH.

use strict;
use warnings;

my %hash1;
   $hash1{hello} = "world!";

foo(\%hash1);
foo( { hello => 'world!!' } );

sub foo
{
  my %hash2 = %{$_[0]};
  print "hello " . $hash2{hello} . "\n";
}
#output
#hello world!
#hello world!!
[download]