this isn't completely possible. no matter what you do with tie, bless, etc. the following lines won't work as you expect: 
$d = "Nov 25, 1971";
print "\$d contains $d\n"; # "Nov 25, 1971"
print "\$d as unixtime:", $d->to_unixtime, "\n";

[download]

with a tie, you can't directly do $d->to_unixtime, it has to be tied($d)->to_unixtime. otherwise you'll get the message: Can't locate object method "to_unixtime" via package "Nov 25, 1971" (perhaps you forgot to load "Nov 25, 1971"?)

with a regular object, you can't of course do $d = "Nov 25, 1971" because this would destroy the object previously assigned to $d.

the most functional hack I could come up with is the use of tie AND overload AND tye's suggestion, as in: 

Datum->tieVar( (my $d), -type=>'date' );

$d = "Nov 25, 1971";

print STDOUT "\$d contains $d\n"; # "Nov 25, 1971"
print STDOUT "\$d as unixtime:", $d->to_unixtime, "\n";

package Datum;

use overload '""' => \&to_string;

sub tieVar {
    my $class= shift @_;
    my $toTie= \$_[0]; shift @_;
    tie $$toTie, $class, @_;
}

sub TIESCALAR {
    my $class = shift;
    my $instance = shift || undef;
    return bless \$instance => $class;
}

sub STORE {
    ${$_[0]} = $_[1];
}

sub FETCH {
    # this just returns "self"
    # (won't be good in some cases)
    return $_[0];
}

sub to_string {
    # this is the "real" FETCH method 
    return ${$_[0]};
}

sub to_unixtime {
    return "foobar";
}

[download]
this prints what you expect: 
$d contains Nov 25, 1971
$d as unixtime:foobar

[download]
...but this is a hack, as it works correctly only when fetching $d in a string context. it has the implicit advantage, however, that $d could return something different (perhaps $d->to_unixtime) in numeric context.

cheers,
Aldo

King of Laziness, Wizard of Impatience, Lord of Hubris

In reply to Re: How might I return tied scalar that does not have to be dereferenced as $$scalar? by dada
in thread How might I return tied scalar that does not have to be dereferenced as $$scalar? by rr

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