I think you're missing the distinction between $a = $b and $a = %b. The first assigns the value of $b to $a, the second sets $a to be a reference to %b. If you to set $a to be a references to $b then you'd do $a = \$b (or possibly $a := $b).

As I understand it, the 'automatic taking of references' thing only applies when you're dealing with an array/hash/sub in a scalar context.

Judging by the work that's going on in parrot, $a = $b won't copy the value of $b immediately, but will create a 'copy on write reference'. A copy will only be made if either of $a or $b is modified.

In reply to Re: Perl6 - value vs. reference issues by pdcawley
in thread Perl6 - value vs. reference issues by John M. Dlugosz

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