Tom Worster wrote:
> On 7/22/09 6:09 PM, "Shawn McKenzie" <[email protected]> wrote:
>
>> Tom Worster wrote:
>>> though the manual is perfectly clear that this should be expected, i was a
>>> bit surprised that the result of the following is 42
>>>
>>> <?php
>>> function foo(&$a) {
>>> $a = 42;
>>> unset($a);
>>> $a = 'meaning';
>>> }
>>> foo($a);
>>> print("$a\n");
>>> ?>
>>>
>>> normally i would expect unset() to free some memory. but in this example it
>>> doesn't and has a different behavior: it releases foo's reference to the
>>> global $a, allowing the next line to define a local $a.
>>>
>>> i think i'd have preferred compile error.
>>>
>>>
>> Well, you unset the reference and then you assigned 'meaning' to a local
>> function variable $a. Why would you get a compile error?
>
> when you state it in those terms (which are clearly correct) i wouldn't.
>
> but if the way i think is "unset() destroys the specified variables" (as the
> manual puts it) then i expect that the specified variable would be
> destroyed, not the reference.
>
> so, as i said, i was a bit surprised when the variable wasn't destroyed.
> once i understood what was happening, i thought it a bit confusing to have
> such scope-dependent differences in behavior of a language element.
>
>
It might be easier to understand if you don't use the same var names:
function foo(&$arg) {
$arg = 42;
unset($arg);
$arg = 'meaning';
}
$a = 0;
foo($a);
print("$a\n");
--
Thanks!
-Shawn
http://www.spidean.com
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