register_shutdown_function converting references to values?

Question

I'm trying to use a shutdown function that uses a value changed dynamically after registration. My understanding was that passing the variable by reference would allow changes in later portions of the code to affect the shutdown call. The code below demonstrates that this is not the case:

<?php
    $x = 0;
    function shutdown(&$x) {echo $x;}
    register_shutdown_function('shutdown',$x);
    $x = 1;
    exit();
?>

Am I misunderstanding how this should be done? Is it expected behavior for the register function to convert the reference pass to a value pass?

PHP version 5.5.9

Answer 1

One of possible solutions is to use anonymous functions (it's a closure in this case with $x captured as a reference):

$x = 0;

$shutdown = function() use(&$x) { echo $x; };

register_shutdown_function($shutdown);
$x = 1;

exit();

Demo: http://ideone.com/L2cYJp

UPD: Explanation why your solution doesn't work:

If you have a look on register_shutdown_function signature:

void register_shutdown_function ( callable $callback [, mixed $parameter [, mixed $... ]] )

you will see that parameters are passed by values. That's it - the value of $x in register_shutdown_function('shutdown',$x); line is passed by value and equals to the value it was on the moment of call.

If doesn't matter if you defined your function that accepts a reference since it's too late to accept a reference because you have already lost reference (indirectly, by using a function that does not take it into account).