Questions about PHP variable separation and reference

Today I read Brother Niao’s article about PHP variable separation and reference. There is a problem that I didn’t understand. I’ll post some screenshots first:

If you follow the above statement, then I will slightly modify the code as follows:

<code><?php
   $var = "laruence";
   $var_dup = &$var;
   $var_ref = &$var;
   $var_ref = "OK";
?>
</code>

Then

Second line of code:
$var_dup and $var point to the same zval with refcount of 2.

When executing the third line:
PHP finds that the refcount of the zval to be operated is greater than 1, then PHP will execute Separation, separate $var_dup, and associate $var and $var_ref with change on write. That is, refcount=2, is_ref=1;

When proceeding to the fourth line:
Since $var and the zval pointed to by $var_ref have is_ref=1;, they will not be separated, making the values ​​​​of $var_ref and $var both "OK".

According to my understanding, at the end of the program, since $var_dup has been separated when executing the third line, its value should remain "laruence" unchanged. However, when I ran the program, I found that it also changed. It became "OK", which makes me very confused. I hope someone who knows the answer can help me. I don’t know if I understood it wrong or if there is another hidden meaning, thank you!

Attached are two small chestnuts for your reference:

<code><?php
   $var = "laruence";
   $var_ref = "OK";
   $var_dup = &$var;
   $var = &$var_ref;


   echo $var;        //OK
   echo $var_dup;    //laruence
   echo $var_ref;    //OK      
?>
</code>

<code><?php
   $var = "laruence";
   $var_ref = "OK";
   $var_dup = &$var;
   $var_ref = &$var;


   echo $var;        //laruence
   echo $var_dup;    //laruence
   echo $var_ref;    //laruence      
?>
</code>

Reply content:

Today I read Brother Niao’s article about PHP variable separation and reference. There is a problem that I didn’t understand. I’ll post some screenshots first:

If you follow the above statement, then I will slightly modify the code as follows:

<code><?php
   $var = "laruence";
   $var_dup = &$var;
   $var_ref = &$var;
   $var_ref = "OK";
?>
</code>

Then

Second line of code:
$var_dup and $var point to the same zval with refcount of 2.

When executing the third line:
PHP finds that the refcount of the zval to be operated is greater than 1, then PHP will execute Separation, separate $var_dup, and associate $var and $var_ref with change on write. That is, refcount=2, is_ref=1;

When proceeding to the fourth line:
Since the is_ref=1; of the zval pointed to by $var and $var_ref, they will not be separated, making the values ​​​​of $var_ref and $var both "OK".

According to my understanding, at the end of the program, since $var_dup has been separated when executing the third line, its value should remain "laruence" unchanged. However, when I ran the program, I found that it also changed. It became "OK", which makes me very confused. I hope someone who knows the answer can help me. I don’t know if I understood it wrong or if there is another hidden meaning, thank you!

Attached are two small chestnuts for your reference:

<code><?php
   $var = "laruence";
   $var_ref = "OK";
   $var_dup = &$var;
   $var = &$var_ref;


   echo $var;        //OK
   echo $var_dup;    //laruence
   echo $var_ref;    //OK      
?>
</code>

<code><?php
   $var = "laruence";
   $var_ref = "OK";
   $var_dup = &$var;
   $var_ref = &$var;


   echo $var;        //laruence
   echo $var_dup;    //laruence
   echo $var_ref;    //laruence      
?>
</code>

Copy On Write! ! ! ! ! ! ! ! ! ! ! !
Copy as you write! ! ! ! ! ! ! ! ! ! ! !

Is there a write operation in the third line of your code? There is no need to perform separation! ?
The third line of operation is to increase refcount to 3.