On Thu, Aug 04, 2016 at 09:36:18PM +0100, Zefram wrote: > Yeah. Let me try to make it clearer. In the above situation, with > a reference to $a's Scalar container in $b, I'd like to achieve what > the assignment "$a = 5" would, but by an operation using $b and not > directly mentioning $a. The result should be that $b is unchanged > (still referencing $a's container) and $a subsequently has the value 5.
So are you looking for...? my $a = 3; my $b := $a; # $b is bound to $a $b = 5; # Assign 5 to $b (which is also $a) say $a; # "5" Pm
