Daniel Convissor wrote:
Hi Gary:
On Thu, Nov 22, 2007 at 08:34:02AM -0500, Gary Mort wrote:
So, in the pass by reference talk, it was pointed out that if you do:
$a = 100;
$b = 100;
// At this point in time, $a and $b are still using the same memory to
store their data
Nope. They have two separate memory assignments. But they'll be the
same if you do the following:
$a = 100;
$b = $a;
Opps, right, sorry about the typo.
Not relevant to my main question, but it does add confusion to have put
a bad example up front.
Thanks.
-Gary
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