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" }