PHP and references are somewhat unintuitive. If used appropriately references in the right places can provide large performance improvements or avoid very ugly workarounds and unusual code.
The following will produce an error:
function f(&$v){$v = true;}
f(&$v);
function f($v){$v = true;}
f(&$v);
None of these have to fail as they could follow the rules below but have no doubt been removed or disabled to prevent a lot of legacy confusion.
If they did work, both involve a redundant conversion to reference and the second also involves a redundant conversion back to a scoped contained variable.
The second one used to be possible allowing a reference to be passed to code that wasn't intended to work with references. This is extremely ugly for maintainability.
This will do nothing:
function f($v){$v = true;}
$r = &$v;
f($r);
More specifically, it turns the reference back into a normal variable as you have not asked for a reference.
This will work:
function f(&$v){$v = true;}
f($v);
This sees that you are passing a non-reference but want a reference so turns it into a reference.
What this means is that you can't pass a reference to a function where a reference is not explicitly asked for making it one of the few areas where PHP is strict on passing types or in this case more of a meta type.
If you need more dynamic behaviour this will work:
function f(&$v){$v = true;}
$v = array(false,false,false);
$r = &$v[1];
f($r);
Here it sees that you want a reference and already have a reference so leaves it alone. It may also chain the reference but I doubt this.