PHP 调用时传递引用是不可避免的吗？

Question

给定以下界面:

interface ISoapInterface {
  public static function registerSoapTypes( &$wsdl );
  public static function registerSoapOperations( &$server );
}

以及以下代码:

$soapProvider = array( "FilePool", "UserList" );
foreach( $soapProvider as $provider ) {
  call_user_func( array( $provider, "registerSoapTypes" ), &$server->wsdl );
  call_user_func( array( $provider, "registerSoapOperations" ), &$server );
}

FilePool 和 UserList 都实现了 ISoapInterface。

PHP 会提示 foreach 中的两个调用:

Call-time pass-by-reference has been deprecated

所以我查看了该消息，the documentation似乎很清楚如何解决这个问题。从实际调用中删除与号。
所以我将代码更改为如下所示:

$soapProvider = array( "FilePool", "UserList" );
foreach( $soapProvider as $provider ) {
  call_user_func( array( $provider, "registerSoapTypes" ), $server->wsdl );
  call_user_func( array( $provider, "registerSoapOperations" ), $server );
}

现在 PHP 报错了

Parameter 1 to FilePool::registerSoapTypes expected to be reference, value given
Parameter 1 to FilePool::registerSoapOperations expected to be reference, value given

除此之外，该功能现已损坏。所以这显然不是解决方案。

Answer 1

来自call_user_func :

Note that the parameters for call_user_func() are not passed by reference.

要调用静态方法，您可以使用 Class::method() 语法，为 Class 和/或 method 部分提供变量:

$soapProvider = array( "FilePool", "UserList" );
foreach( $soapProvider as $provider ) {
  $provider::registerSoapTypes($server->wsdl);
  $provider::registerSoapOperations($server);
}