On Mon, Apr 30, 2001 at 11:17:06AM -0700, Jay Strauss wrote:
> Lets say I have a database handle object, $dbh. Now I
> want to pass it to another subroutine like:
>
> backup(dbh=>$dbh);
>
> sub backup
> {
> my (%arg) = @_;
> my $sth = $arg{"dbh"}->prepare(some SQL);
> my $rc = $sth->execute();
> }
>
> Now this works, but shouldn't I be passing a reference
> to the object, and then unreferencing it in the sub
> routine?
You already are!
>
> I tried:
>
> backup (dbh=>\$dbh);
>
> sub backup
> {
> my (%arg) = @_;
> my $dbh = \$arg{"dbh"};
Hm... $arg{"dbh"} is *already* a reference to a reference to an
object, so applying the \ operator makes it a reference to a reference
to a reference of an object.
I think $dbh = ${$arg{"dbh"}} would've worked, but this is pretty pointless.
> my $sth = $dbh->prepare(some SQL);
> my $rc = $sth->execute();
> }
>
> But that didn't work. Maybe this is all stupid
> because I think $dbh is just a pointer to an object,
> and so passing by value just gives me another pointer
> to an object.
Pretty much, yeah - except that Perl doesn't have pointers. But if
you meant reference, that's absolutely correct.
In fact, this is always the case, since Perl's definition of an object
is a reference (any kind, but usually a hashref) that has been
"blessed" into a class.
