Question

Variable scope (as defined here)

The scope of a variable is the context within which it is defined. For the most part all PHP variables only have a single scope. This single scope spans included and required files as well.

//a.php
<?php
class a {
 function &func () {
   $avar = array("one", "two", "three");
   return $avar;
}
?>

__

//b.php
<?php
class b {
 include("a.php");
 $ainstance = new a;
 var_dump($ainstance->func()); 
}
?>

The above code will Dump information about the variable as expected (I mean WRT the structure as formed in the function func).

My doubt is that,

• where are the variable stored when it is in the scope of a function?
• If it is on the call stack, then wont the variable be cleaned/destroyed when the function terminates?
• Since the variable is not getting destroyed (as per the code above), why is it not getting destroyed or does PHP have a mechanism to save the variable (say in a heap)and return the reference to it.
• Does PHP have call stack at all?

Answer 1

All variables in PHP are zval*, that is, C pointers.

If you return by value, PHP will, in most cases, automatically copy the zval* and return that. If you return by reference, PHP will return the original zval*. In none of these cases does the returned zval* refcount reaches 0.

On the C side, well, when you return a variable, it returns a pointer to a zval, which is a C struct containing information about the variable (namely type, value, the refcount and a is_ref flag).

Since it’s a pointer, it’s not actually returning a local C variable, but a pre-allocated zval pointer, which points to the location of the actual zval. Unless that zval* refcount reaches 0 (i.e.: not storing the return value), the variable will still live until the end of the program.