in reply to Perl6 - value vs. reference issues
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.
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Re: Re: Perl6 - value vs. reference issues
by jryan (Vicar) on Sep 14, 2002 at 19:30 UTC |