On Mon, Jun 27, 2011 at 2:44 PM, <[email protected]> wrote:
>
> Found the solution (my apologies).
>
> I am receiving both an scalar, the second one as a reference, then it must
> be assigned to an other var.
>
> No. It does not have to be assigned to another var. Instead you should
dereference the reference.
sub dosomething
> {
> ($myopt,$myparams) = @_;
> ## %myparams = $myparams; # skip the asssignment and creation oa a
> new hash
> print "opt = $myopt\n";
>
#### while( my ($k, $v) = each %myparams )
while( my ($k, $v) = each %{$myparams} ) # dereference the
reference with %{$ref}
> { print "$k = $v \n"; }
> }
>
>
The assignment you had assigns the hash reference to a key of the hash you
created, with nothing assigned as a value.
It is a real, real bad idea (for many reasons) to create variables of
different types with the same names.
A major benefit of passing a reference is that you only move one item, the
reference, instead of the entire hash. It doesn't make a lot of difference
in this case, but if you had a hash with thousands of key/value pairs
passing a reference and using that reference uses just a few bytes of
memory. Copying the entire hash uses many thousands of bytes.
Happy hacking,
Mike
