Posted by Vince Morgan on 05/08/07 00:50

"Jerry Stuckle" <jstucklex@attglobal.net> wrote in message
news:kpqdnd8GMfhP36LbnZ2dnUVZ_rOqnZ2d@comcast.com...
> 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>
>
Thank you for taking the time Jerry.
I haven't the time just yet to read the entire linked article, but have read
enough already to realize my understanding of references in php is
incorrect.
Thanks again,
Vince

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