On Mon, Jun 27, 2011 at 2:44 PM, 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.
su
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.
sub dosomething
{
($myopt,$myparams) = @_;
%myparams = $myparams;
print "opt = $myopt\n";
while( my ($k, $v) = each %myparams )
