Are multiple variable assignments done by value or reference?

Question

$a = $b = 0; 

In the above code, are both $a and $b assigned the value of 0, or is $a just referencing $b?

Answer 1

With raw types this is a copy.

test.php

$a = $b = 0;  $b = 3;   var_dump($a); var_dump($b); 

Output:

int(0)  int(3) 

With objects though, that is another story (PHP 5)

test.php

class Obj {      public $_name; }  $a = $b = new Obj();  $b->_name = 'steve';  var_dump($a); var_dump($b); 

Output

object(Obj)#1 (1) { ["_name"]=> string(5) "steve" }  object(Obj)#1 (1) { ["_name"]=> string(5) "steve" } 

Answer 2

Regard this code as:

$a = ($b = 0); 

The expression $b = 0 not only assigns 0 to $b, but it yields a result as well. That result is the right part of the assignment, or simply the value that $b got assigned to.

So, $a gets assigned 0 as well.