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