Katharina Wolkwitz wrote:
> Hi again!
> 
> I think I've found the solution to my problem:
> 
> Changing the following line in accesscontrol.php from:
> 
> function doControlUserAccess( $input, $argv, &$parser )

prototype means the third parameter is a reference.

Since it was given a value, per the error message, changing the
prototype to

> function doControlUserAccess( $input, $argv, $parser )

fixes it.


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