$a = new stdClass;
$b = $a;
$a->foo = 'bar';
/* Notice at this point, that $a and $b are,
* indeed sharing the same object instance.
* This is their reference-like behavior at work.
$a = 'baz';
/* Notice now, that $b is still that original object.
* Had it been an actual reference with $a,
* it would have changed to a simple string as well.