> -----Original Message-----
> From: Randal L. Schwartz [mailto:[EMAIL PROTECTED]
> Sent: Wednesday, December 24, 2003 3:50 PM
> To: [EMAIL PROTECTED]
> Subject: Re: the ref() function: what does it mean when ...
>
>
> >>>>> "Wiggins" == Wiggins D Anconia <[EMAIL PROTECTED]> writes:
>
> Wiggins> Out of curiousity, why/when *in Perl* would you take a
> Wiggins> reference to something that holds a reference? And,
> "how deep
> Wiggins> does the well go?" (how far will Perl take the above
> Wiggins> indirection?)... I suppose I could just test, but I
> am rather
> Wiggins> lazy...
>
> For the same reason you might take a reference to a scalar
> otherwise... indirection.
>
> For example, I might have
>
> my $active_table = [EMAIL PROTECTED];
>
> and then I want to call a subroutine to decide a new active table:
>
> adjust_table(\$active_table);
>
> sub adjust_table {
> my $table_ref = shift;
> $$table_ref = rand(2) > 1 ? [EMAIL PROTECTED] : [EMAIL PROTECTED];
> }
>
> I can't change $active_table except by reference.
>
My question would be, why force the reference issue at all when
you could do something like:
my $active_table = [EMAIL PROTECTED];
$active_table=adjust_table();
sub adjust_table {
return(rand(2) > 1 ? [EMAIL PROTECTED] : [EMAIL PROTECTED]);
}
Shawn
--
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
<http://learn.perl.org/> <http://learn.perl.org/first-response>
