PHP中的數組引用混淆

Question

$arr = array(1); 
$a = & $arr[0]; 

$arr2 = $arr; 
$arr2[0]++; 

echo $arr[0],$arr2[0]; 

// Output 2,2 

你能幫我一下嗎？怎麼可能？

Answer 1

但是，請注意，數組內部的引用可能是危險的 。使用右側的 引用進行正常（非引用）分配不會將左側變爲 引用，但在這些正常的 賦值中將保留數組內的引用。這也適用於數組傳遞值爲 的函數調用。

/* Assignment of array variables */ 
$arr = array(1); 
$a =& $arr[0]; //$a and $arr[0] are in the same reference set 
$arr2 = $arr; //not an assignment-by-reference! 
$arr2[0]++; 
/* $a == 2, $arr == array(2) */ 
/* The contents of $arr are changed even though it's not a reference! */ 

Answer 2

它看起來像$改編[0]和$ ARR2 [0]都指向同一個分配的內存，所以如果你增加的指針之一，INT引腳將在存儲

鏈接Are there pointers in php?被遞增

Answer 3

$arr = array(1);//creates an Array ([0] => 1) and assigns it to $arr 
$a = & $arr[0];//assigns by reference $arr[0] to $a and thus $a is a reference of $arr[0]. 
//Here $arr[0] is also replaced with the reference to the actual value i.e. 1 

$arr2 = $arr;//assigns $arr to $arr2 

$arr2[0]++;//increments the referenced value by one 

echo $arr[0],$arr2[0];//As both $aar[0] and $arr2[0] are referencing the same block of memory so both echo 2 

// Output 22