note 73459 added to language.references.return

The note by zayfod below is very misleading. The following is not true: "You do not have to use & to indicate that reference binding should be done when you assign to a value passed by reference the result of a function which returns by reference." As proof, you can change zayfod's second example slightly to show that $var does not, as he claims, become a reference to $some_var. <?php class some_class { function & func_b () { $this->some_var = 2; return $this->some_var; } function func_a (& $param) { # $param is 1 here $param = $this->func_b(); # $param is 2 here } } $var = 1; $object = new some_class(); $object->func_a($var); print $var; // this prints two $var = 5; print $object->some_var; // $object->some_var is still 2, because $var is not bound to it // (as the original post by zayfod claims) and therefore has no // effect on it ?> In actuality it is impossible to bind a variable passed by reference to a variable returned by reference, doing so only removes the link between the internal parameter variable and binds it to something new. This is why zayfod's first example shows $var having a value of 1 at the end. What is actually happening in zayfod's first example is the following: 1) $var is passed by reference into func_a, which binds $param to $var 2) func_b is evaluated, returning a reference to $some_var 3) $param is bounded to $some_var therefore losing it's binding to $var 4) func_a returns, with $var unchanged In the second example step 1 and 2 are the same, but then this happens: 3) the value 2 is copied into $param, which still references $var, and therefore $var now also evaluates to 2 4) func_a returns, $var is 2, but is not bound to $some_var (and wouldn't be even in my example where some_var still exists afterward) The mistake seems to be based on misuse of references. When you return a reference, it means that you want a variable that is tied to another, so that when you change one, the other one changes. In zayfod's example, returning by reference is pointless, because $some_var ceases to exist after the function is completed, it won't stay tied to anything. There are two ways to change those functions so that in the end you have $var actually bound to $this->some_var: 1) Pass $param into func_b by reference and reference $this->some_var to func_b's parameter like so: func_b (&$b_param) { $this->some_var =& $b_param; // be careful not to flip this $this->some_var = 2; } 2) change func_a to return the result of func_b by reference, and reference that to $var And remember this simple rule: If a variable is already referenced to another variable and you reference it to a third (put it on the left side of =&) it loses its original reference. This is what happens to $param in zayfod's example. (Also don't forget that passing by reference actually references the internal parameter variable to the the passed variable.) ---- Server IP: 64.71.164.2 Probable Submitter: 12.197.12.50 ---- Manual Page -- http://www.php.net/manual/en/language.references.return.php Edit -- https://master.php.net/note/edit/73459 Del: integrated -- https://master.php.net/note/delete/73459/integrated Del: useless -- https://master.php.net/note/delete/73459/useless Del: bad code -- https://master.php.net/note/delete/73459/bad+code Del: spam -- https://master.php.net/note/delete/73459/spam Del: non-english -- https://master.php.net/note/delete/73459/non-english Del: in docs -- https://master.php.net/note/delete/73459/in+docs Del: other reasons-- https://master.php.net/note/delete/73459 Reject -- https://master.php.net/note/reject/73459 Search -- https://master.php.net/manage/user-notes.php