On Aug 19, 11:16 am, [EMAIL PROTECTED] (Perry Smith) wrote:
> My friend is trying to do this:
>
> #!/bin/perl
>
> $a = 1;
> $b = 2;
> $c = 3;
>
> @l = ( \$a, \$b, \$c );
>
> # This is the line we need help with
> # Is there a way to tell Perl that @l[0] is a reference
> # and we want to assign what it refers to the number 4.
> # i.e. ${$l[0]} = 4;
> # but for the whole list in @l
> ( \( @l ) ) = ( 4, 5, 6 );
>
> print "$a should be 4\n";
It sounds to me like you're looking for the exact opposite of the \
( ... ) behavior. That is, you want some operation that will take a
list of references and return the corresponding list of referents.
To my knowledge, no such operator exists. Sorry.
You can dereference a single reference to an array. But you can't
dereference a list of references to scalars.
Best you could do would probably be something involving a postfix for,
like:
my @new_vals = (4, 5, 6);
my $i = 0;
${$_} = $new_vals[$i++] for @l;
Paul Lalli
--
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
http://learn.perl.org/
