>Ahh, but that's very different. My suspicion is that both the last-expr >scalar of the block and the $result scalar will share the same single >payload, similar to how:
> $x = $y >doesn't actually copy the *content*... just the scalar wrapper around >the content. Does this mean that $x, $y taken as references internal by Perl interpreter actually? $x = $y, means that $x, $y shares the same memory but not allocate new memory for $x then copy the content of $y to $x ? -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
