Another example of something to watch out for when using references with arrays.  It seems that even an usused reference to an array cell modifies the *source* of the reference.  Strange behavior for an assignment statement (is this why I've seen it written as an =& operator?  - although this doesn't happen with regular variables).
<?php
    $array1 = array(1,2);
    $x = &$array1[1];   // Unused reference
    $array2 = $array1;  // reference now also applies to $array2 !
    $array2[1]=22;      // (changing [0] will not affect $array1)
    print_r($array1);
?>
Produces:
    Array
    (
    [0] => 1
    [1] => 22    // var_dump() will show the & here
    )

//above was Noted By Dave at SymmetricDesign dot com//
//and below is my opinion to this simple problem. //

This is an normal referencing problem.

when you gain an reference to a memory at some variable.
this variable, means "memory itself". (in above example, this would be ->  $x = &$array1[1];   // Unused reference)

and you've copied original one($array1) to another one($array2).
and the copy means "paste everything on itself". including references or pointers, etcs. so, when you copied $array1 to $array2, this $array2 has same referencers that original $array1 has. meaning that $x = &$array1[1]  = &$array2[1];
and again i said above. this reference means "memory itself".
when you choose to inserting some values to $array2[1], 
$x; reference is affected by $array2[1]'s value. because, in allocated memory, $array2[1] is a copy of $array1[1]. this means that $array2[1] = $array1[1], also means &$array2[1] = &$array1[1]  as said above. this causes memory's  value reallocation on $array1[1]. at this moment. the problem of this topic is cleared by '$x', the memory itself.  and this problem was solved by unsetting the '$x'. unsetting this reference triggers memory reallocation of $array2[1]. this closes the reference link between the copied one($array1, which is the original) and copy($array2). this is where that bug(clearly, it's not a bug. it's just a missunderstanding) has triggered by. closing reference link makes two array object to be separated on memory. and this work was done through the unset() function. this topic was posted 7 years ago, but i just want to clarify that it's not a bug.

if there's some problems in my notes, plz, note that on above.