Passing/Returning 引用对象 + 更改对象无效

Passing/Returning references to object + changing object is not working

我正在使用 an answer from How to get random value out of an array to write a function that returns a random item from the array. I modified it to pass by reference and return a reference

很遗憾,它似乎不起作用。对返回对象的任何修改都不会保留。我做错了什么?

我正在使用 PHP 5.4,如果有影响的话(不要问)。

function &random_value(&$array, $default=null)
{
    $k = mt_rand(0, count($array) - 1);
    $return = isset($array[$k])? $array[$k]: $default;
    return $return;
}

用法...

$companies = array();
$companies[] = array("name" => "Acme Co",  "employees"=> array( "John", "Jane" ));
$companies[] = array("name" => "Inotech",  "employees"=> array( "Bill", "Michael" ));

$x = &random_value($companies);
$x["employees"][] = "Donald";
var_dump($companies);

输出...

array(2) {
  [0] =>
  array(2) {
    'name' =>
    string(7) "Acme Co"
    'employees' =>
    array(2) {
      [0] =>
      string(4) "John"
      [1] =>
      string(4) "Jane"
    }
  }
  [1] =>
  array(2) {
    'name' =>
    string(7) "Inotech"
    'employees' =>
    array(2) {
      [0] =>
      string(4) "Bill"
      [1] =>
      string(7) "Michael"
    }
  }
}

我什至复制并粘贴了文档中的示例,并且 none 中的示例都有效。他们都输出null.

三元运算符创建一个隐式副本,这会破坏引用链。使用明确的 if... else:

function &random_value(&$array, $default=null)
{
    $k = mt_rand(0, count($array) - 1);
    if (isset($array[$k])) {
        return $array[$k];
    } else {
        return $default;
    }
}

至于为什么,docs 现在状态:

Note: Please note that the ternary operator is an expression, and that it doesn't evaluate to a variable, but to the result of an expression. This is important to know if you want to return a variable by reference. The statement return $var == 42 ? $a : $b; in a return-by-reference function will therefore not work and a warning is issued in later PHP versions.

另见 this bug 三元运算符实际上 returns 在 foreach 上下文中引用,而实际上不应该这样做。

由于我刚刚吃了一顿丰盛的午餐,所以我无法思考比@bishop 更好的功能方面,但是这有效:

$x =& $companies[array_rand($companies)];
$x["employees"][] = "Donald";
var_dump($companies);