note 102593 modified in language.references.return by danbrown

Keep in mind that returning by reference doesn't work with __callStatic: <?php class Test { private static $_inst; public static function & __callStatic ($name, $args) { if (!isset(static::$_inst)){ echo "create"; static::$_inst = (object)"test"; } return static::$_inst; } var_dump($a = &Test::abc()); // prints 'create' $a = null; var_dump(Test::abc()); // doesn't prints and the instance still exists in Test::$_inst ?> --was-- Keep in mind that returning by reference doesn't work with __callStatic: class Test { private static $_inst; public static function & __callStatic ($name, $args) { if (!isset(static::$_inst)){ echo "create"; static::$_inst = (object)"test"; } return static::$_inst; } var_dump($a = &Test::abc()); // prints 'create' $a = null; var_dump(Test::abc()); // doesn't prints and the instance still exists in Test::$_inst http://php.net/manual/en/language.references.return.php