On Fri, 2009-05-22 at 07:44 -0500, Shawn McKenzie wrote: > kranthi wrote: > > i have this script > > > > <?php > > $x = 1; > > $y = 2; > > $a1 = array(&$x, &$y); > > $a2 = array($x, $y); > > $a2[0] = 3; > > print_r($a1); > > print_r($a2); > > ?> > > > > i am expecting > > > > Array > > ( > > [0] => 3 > > [1] => 2 > > ) > > Array > > ( > > [0] => 3 > > [1] => 2 > > ) > > > > > > while i m getting > > > > Array > > ( > > [0] => 1 > > [1] => 2 > > ) > > Array > > ( > > [0] => 3 > > [1] => 2 > > ) > > > > > > any ideas why this is happening?? or am i missing something..? > > the same is the case when i replace > > $a2[0] = 3; with > > $a1[0] = 3; > > $x = 3; > > > > Kranthi. > > > > $a2[0] was assigned the value of $x or 1, so when you change $a2[0], > that's all that changes. You have changed the value 1 to 3. $a1[0] is a > reference to $x, so if you change $a1[0] it will change $x, but not > $a2[0] because it is not a reference to $x. > > -- > Thanks! > -Shawn > http://www.spidean.com > To get the results you expect, you should remove the & from the assignment, as this is assigning by reference, rather than by value like Shawn said, or use an & for $a2 as well. Ash www.ashleysheridan.co.uk -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
