Re: [vox-tech] A question of perl style

[Dale Bewley <MAPSdale@bewley.net>]

yea if you just pass in $dbh you are passing in a reference. and it is not
a big mem copy or anything inefficient like passing in a whole array
instead of a ref to an array or something.

you can always do a ref() on something to find out what it is.

[dale@eng database]$ perldoc -f ref
=item ref EXPR

=item ref

Returns a TRUE value if EXPR is a reference, FALSE otherwise.  If EXPR
is not specified, C<$_> will be used.  The value returned depends on the
type of thing the reference is a reference to.
Builtin types include:

    REF
    SCALAR
    ARRAY
    HASH
    CODE
    GLOB

If the referenced object has been blessed into a package, then that
package
name is returned instead.  You can think of C<ref()> as a C<typeof()>
operator.

    if (ref($r) eq "HASH") {
        print "r is a reference to a hash.\n";
    }
    unless (ref($r)) {
        print "r is not a reference at all.\n";
    }
    if (UNIVERSAL::isa($r, "HASH")) {  # for subclassing
        print "r is a reference to something that isa hash.\n";
    } 

See also L<perlref>.

On Mon, 30 Apr 2001, 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?
> 
> I tried:
> 
> backup (dbh=>\$dbh);
> 
> sub backup
> {
>    my (%arg) = @_;
>    my $dbh = \$arg{"dbh"};
>    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.
> 
> Jay
> 
> 
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--
Dale Bewley - Bewley Internet Solutions Inc. http://bewley.net/