Why the refcount is 2 not 1?

Question

  $var = 1;
  debug_zval_dump($var);

Output:

long(1) refcount(2)


  $var = 1;
  $var_dup = &$var;
  debug_zval_dump($var);exit         


        

Answer 1

I'll try to give some more light to the debug_zval_dump() function and the way you process your variables. Don't kill me if I'm wrong :)...

  $var = 1;
  debug_zval_dump($var);

I think the debug function counts the $var refcount(1) and the 1 refcount(2) since 1 is the value of $var.
If you look at it logically you are actually saying this.

  1 = 1;
  debug_zval_dump(1);

Second part:

$var = 1;
$var_dup = &$var;
debug_zval_dump($var);exit;

What you see here is that you set $var to $var_dup but is keeping its value. The refcount of $var is 1 because you 'linked' it to $var_dup.

$var = 2;
$var_dup = &$var; //or $var = &$var_dup; (doesn't matter which one)
$var = 3;
debug_zval_dump($var_dup);exit;

This gives long(3) refcount(1)... Why is it refcount 1? As you can see the value of $var_dup was never assigned to 3, it should be 2 right? No it shouldn't because you keep it up to date with &$var. This means that when you past $var = 4 between $var = 3 and debug_zval_dump($var_dup);exit; the value of $var_dup will be updated automatically because you have linked them, making it 1 refcount.

Then there is this other occurrence:

$var = 2;
$var_dup = $var;
$var = 4;
debug_zval_dump($var_dup);exit;

The output of this is: long(2) refcount(2). As you can see the value of $var_dup is correct. $var was 2, the value was passed through $var_dup an he sticked with it. The refcount is 2 because is counts $var = 4; and $var_dup = $var;. When we remove the $var = 4; we get this:

$var = 2;
$var_dup = $var;
debug_zval_dump($var_dup);exit;

The output of this is: long(2) refcount(3). Now the debug function count the following: $var_dup(1), =$var(2) (since $var_dup was originated from $var) and $var(= 2;)(3).

I'll hope you understand what I mean. In my opinion this is more math then programming, so that may be the reason why it is a difficult function to understand.

And again, if I'm wrong, don't kill me :)...
Greetings,
Mixxiphoid

Disclaimer
I do not know what the purpose is of this function. I actually never heard of it until today. So I'm not responsible for inappropriate use :).