«

Hi

In the following sequence

$r = {}; # a hashref
$r = []; # an arrayref
$r = ?; # a scalar ref

What is ? ? I want to pass a ref to a scalar (pass by reference)
without resorting to

my $r;
mySub( \$r );

I just want to use

my $r = (something);
mySub( $r );

to be consistent with

my $r = {}; # or my $r = [];
mySub( $r );

I know it's not a big drama on the surface, but I'm trying to overload
the method to return various results depending on the reference type,
and I dont want the \ in some calls and not others.

Thanks

>>>>> "CB" == Chet Butcher <> writes:

CB> In the following sequence

CB> $r = {}; # a hashref
CB> $r = []; # an arrayref
CB> $r = ?; # a scalar ref

CB> What is ? ? I want to pass a ref to a scalar (pass by reference)
CB> without resorting to

there is no special syntax to get a scalar ref like with hashes and arrays.

CB> my $r;
CB> mySub( \$r );

a do block works fine too:

my $r = do{ \my $r } ;

CB> I know it's not a big drama on the surface, but I'm trying to overload
CB> the method to return various results depending on the reference type,
CB> and I dont want the \ in some calls and not others.

you need the \ somewhere to generate a scalar ref. and like with arrays
and hashes you can get them in loops and subs without the do block:

<untested pseudo code>

while( 1 )

my $foo = get_stuff() ;
push @foos, \$foo ;
}

perl will allocate a fresh scalar so you get a fresh scalar ref each
iteration. this is like:

while( 1 )

my @bar = get_list() ;
my %baz = get_hash() ;


push @stuff, { bar => \@bar, baz => \%baz } ;
}

you get new arrays and hashes each iteration because a ref to the old
value is still around (inside @stuff's hashes).

uri

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There's not a special syntax for scalar refs, as there is for array refs
and hash refs. You can do this:

my $r = \"abc";
my $r = \undef;

etc...

Chet Butcher wrote:

> In the following sequence
>
> $r = {}; # a hashref
> $r = []; # an arrayref
> $r = ?; # a scalar ref
>
> What is ? ? I want to pass a ref to a scalar (pass by reference)

$ perl -wle'
my $v = 1;
my $r = \$v;
$$r = 2;
print $v;
'
2

--
Ruud

On Fri, 30 Jan 2009 22:47:53 +0000, Chet Butcher <> wrote:

>Hi
>
>In the following sequence
>
>$r = {}; # a hashref
>$r = []; # an arrayref
>$r = ?; # a scalar ref
>
>What is ? ? I want to pass a ref to a scalar (pass by reference)
>without resorting to
>
> my $r;
> mySub( \$r );
>
>I just want to use
>
> my $r = (something);
> mySub( $r );
>
>to be consistent with
>
> my $r = {}; # or my $r = [];
> mySub( $r );
>
>I know it's not a big drama on the surface, but I'm trying to overload
>the method to return various results depending on the reference type,
>and I dont want the \ in some calls and not others.
>
>Thanks

Not sure that overloads can be done in Perl, maybe.
Wether the method expects a reference, and what type of reference, or not, is up to you.
Use alias parameter processing, calls can be general, figure out specifics in the method.
Then you could return success while processing data directly (one schema - are many more).

sln

-------------------------------------
sub ProcessData
{
return 0 if (@_ < 1);
my $Dataref;
if (!length( ref($_[0]) ) {
$Dataref = \$_[0];
} else {$Dataref = $_[0]}
shift;

if (ref($Dataref) eq 'SCALAR') {
return ProcessScalar($Dataref, @_);
}
if (ref($Dataref) eq 'ARRAY') {
return ProcessArray($Dataref, @_);
}
if (ref($Dataref) eq 'HASH') {
return ProcessHash($Dataref, @_);
}
return 0;
}

sub ProcessScalar
{
my ($scalar_ref, ...) = @_;
}
sub ProcessArray
{
my ($array_ref, ...) = @_;
}
sub ProcessHash
{
my ($hash_ref, ...) = @_;
}