Hi, I falied in passing by reference using call_user_function().
In case 2 and 5, I expect that $a and $b are overwritten by MyMethod or MyFunction. However, they were not overwritten.
Is this right doing? Or is my understanding wrong?

Regards,

KUBO Atsuhiro


test code:
<?php
class MyClass {
    function MyMethod(&$arg1, &$arg2) {
        $arg1 = 'foo';
        $arg2 = 'bar';
    }
}

print "case 1:\n";
$a = 1;
$b = 2;
print '$a = ' . $a . "\n";
print '$b = ' . $b . "\n";

print "case 2:\n";
$a = 1;
$b = 2;
$obj = new MyClass;
call_user_func(array(&$obj, 'MyMethod'), $a, $b);
print '$a = ' . $a . "\n";
print '$b = ' . $b . "\n";

print "case 3:\n";
$a = 1;
$b = 2;
call_user_func(array(&$obj, 'MyMethod'), &$a, &$b);
print '$a = ' . $a . "\n";
print '$b = ' . $b . "\n";

print "case 4:\n";
$a = 1;
$b = 2;
MyFunction($a, $b);
print '$a = ' . $a . "\n";
print '$b = ' . $b . "\n";

print "case 5:\n";
$a = 1;
$b = 2;
call_user_func('MyFunction', $a, $b);
print '$a = ' . $a . "\n";
print '$b = ' . $b . "\n";

print "case 6:\n";
$a = 1;
$b = 2;
call_user_func('MyFunction', &$a, &$b);
print '$a = ' . $a . "\n";
print '$b = ' . $b . "\n";

exit();

function MyFunction(&$arg1, &$arg2) {
    $arg1 = '1st';
    $arg2 = '2nd';
}
?>

output:
$a = 1
$b = 2
case 2:
$a = 1
$b = 2
case 3:
$a = foo
$b = bar
case 4:
$a = hoge
$b = huge
case 5:
$a = 1
$b = 2
case 6:
$a = hoge
$b = huge
?>