Question

我们有以下代码：

$fruits = array(
    (object) array(
        'name' => 'orange',
        'country' => 'brazil',
        'qty' => 23,
        'destination' => 'romania'
    ),
    (object) array(
        'name' => 'pinapple',
        'country' => 'thailand',
        'qty' => 3,
        'destination' => 'germany'
    ),
    (object) array(
        'name' => 'carrot',
        'country' => 'romania',
        'qty' => 250,
        'destination' => 'france'
    ),
    (object) array(
        'name' => 'bannana',
        'country' => 'kenya',
        'qty' => 50,
        'destination' => 'sweden'
    ),
    (object) array(
        'name' => 'melon',
        'country' => 'romania',
        'qty' => 50,
        'destination' => 'uae'
    )
);

function testing($fruits)
{
    for($i = 0, $size = sizeof($fruits); $i < $size; $i++)
    {
        $fruit = $fruits[$i];
        $fruits[$i]->qty = $fruit->qty * 100;

        unset($fruit->destination);
        unset($fruit->test);
        unset($fruit->country);
    }

    return $fruits;
}

var_dump($fruits);

testing($fruits);

var_dump($fruits);

?>

如果您运行该代码，$ fruits数组的项目已更改，因为它已作为参考传递。

问题是：我们如何传递它以便tests（）函数不会改变$ fruits？

感谢所有人。

Answer 1

如何克隆你的对象？

$clone = array();

foreach ($fruits as $k => $v) {
    $clone[$k] = clone $v;
}
testing($clone);

PHP变量赋值被视为引用

1 个答案: