php - PHP 中的链接引用

Question

我试图理解 PHP 引用，但在使用链接引用时发现了一个问题。

class A
{
    public $val;

    public function __construct($val)
    {
        $this->val = $val;
    }
}

$values = array(
    'a' => new A('a'),
    'b' => new A('b'),
    'c' => new A('c')
);

$values['a'] = &$values['b'];
$values['b'] = &$values['c'];

返回:

array(
'a' => new A('b'),
'b' => new A('c'),
'c' => new A('c')
)

为什么对象 A 中键“a”的“val”是“b”？我预计该值为“c”。谢谢

Answer 1

尽量避免引用，它们可能会给您带来难以调试的麻烦。此外，对象在幕后被引用；这意味着:

$a1 = new A('a');
$a2 = $a1;

类似于:

$b1 = new A('a');
$b2 = &$b1;

顺便说一句:

$a2->val = 1; // $a1->val is now equal to 1, because $a1 and $a2 are pointing 
              // to the same instance
$b2->val = 1; // $b1->val is now equal to 1, because $b2 points to $b1

但是有一个微妙的区别:

$a1 = 1; // $a2 still points to A object
$b1 = 1; // $b2 still points to $a1 which points to number 1,
         // therefore $b2 == 1

此外，它与数组的工作方式略有不同，因为数组赋值总是涉及值复制。

如果您想了解示例中发生的情况，让我们看一下:

所以你的原始数组是这样的:

$values = array(
    'a' => new A('a'),
    'b' => new A('b'),
    'c' => new A('c')
);

让我们一步步看看里面发生了什么:

$values['a'] = &$values['b']; // $values['a'] is now reference to new A('b')
                              // that means your original new A('a') is now
                              // lost

$values['b'] = &$values['c']; // $values['b'] is now reference to new A('c')
                              // stored under 'c' key, that means $values['b']
                              // is now equal to $values['c'] ; note that this is 
                              // different than $b2 = &$b1; from the above example
                              // since we use an array and not bare variables
                              // the $values['a'] points to value stored under 
                              // the 'b' key, but we replace the 'b' key value 
                              // as opposed to giving it a new value; 
                              // "Array assignment always involves value copying."

// So you ended up with this result:
array(
  'a' => new A('b'),
  'b' => new A('c'), // this is actually reference to 'c', just wrote new A() 
                     // to keep this part consistent with your question's
                     // formatting
  'c' => new A('c') 
)