The foreach loop uses a copy of the original array. Why can the original array be changed after reference assignment?

$aOriginArray为二维数组
Array
(
    [0] => Array
        (
            [test1] => 1
            [test2] => 2
        )
)
foreach($aOriginArray as &$aNewValue){
    //为原数组子数组添加一个元素
    $aNewValue['new'] = 3;
}
//经过foreach循环引用赋值方式修改数组后的结果
print_r($aOriginArray);
Array
(
    [0] => Array
        (
            [test1] => 1
            [test2] => 2
            [new]   => 3 
        )
)

This paragraph can prove that the foreach operation is a copy of the original array.
Moreover, the original array was lost halfway during synchronization. The internal pointer of the original array was only moved once.
Subsequent operations are all in

$k = ['a','b','c'];

foreach($k as $t){
    echo "current:".current($k);
    echo "<br>";
    echo "foreach:".$t;
    echo "<br>";
}
输出的结果是:
current:b
foreach:a
current:b
foreach:b
current:b
foreach:c

When using reference assignment, the reference situation of the array is as follows

xdebug_debug_zval("k");

k:
(refcount=2, is_ref=1),

数组的引用计数为2，说明有两个变量指向同一个数组,is_ref值为1,说明数组被引用
也就是说foreach在引用赋值时操作的是原数组,此时没有产生数组的拷贝
foreach的第一个参数为原数组的别名

foreach($k as &$t){
    ...
}

当使用foreach($k as $t){...}不加引用
数组的引用情况如下
k:
(refcount=1, is_ref=0),

原数组仅有一个符号在指向且未被引用

Thank you very much for your patient answers