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Posted by Jerry Stuckle on 05/07/07 16:16
Vince,
Almost - but the assignment into $var is an operation, also. More below.
Vince Morgan wrote:
> "Vince Morgan" <vinharAtHereoptusnet.com.au> wrote in message
> news:463f3216$0$15845$afc38c87@news.optusnet.com.au...
>> <?
>> function &get()
>> {
>> global $r;
>> return $r;
>> }
>> $r=5;
>> $var=$r;
>> echo $var;//output is 5 as expected.
>> $r=6;
>> echo $var;//output is still 5 as not expected, by me that is ;)
>> $var=$r;
>> echo $var;//output is now 6
>> ?>
>
> Oooops, getting too late evidently. I meant to write the following.
> <?
> function &get()
> {
> global $r;
> return $r;
> }
> $r=5;
> $var=get();
get() is returning a reference. But you are assigning a copy into $var.
> echo $var;//output is 5 as expected.
> $r=6;
> echo $var;//output is still 5
> $var=get();
Same as above.
> echo $var;//output is now 6
> ?>
>
>
If you change these to
$var = $get();
You will get 5, 6 and 6, respectively.
There's a discussion about it at
<http://www.php.net/manual/en/language.references.return.php>
--
==================
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Jerry Stuckle
JDS Computer Training Corp.
jstucklex@attglobal.net
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