传递/返回对对象的引用 + 更改对象不起作用 [英] Passing/Returning references to object + changing object is not working

问题描述

我正在使用 一个答案来自如何从数组中获取随机值 编写一个从数组中返回随机项的函数.我将其修改为 pass by reference 和 返回引用.

I am using 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.

不幸的是，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?

Unfortunately, it doesn't appear to work. Any modifications to the returned object do not persist. What am I doing wrong?

我使用的是 PHP 5.4(不要问).

I'm on PHP 5.4 if that makes a difference (Don't ask).

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，但这些示例都不起作用.他们都输出null.

I even copied and pasted the examples from the documentation and none of those worked either. They all output null.

推荐答案

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

The ternary operator creates an implicit copy, which breaks the reference chain. Use an explicit 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;
    }
}

至于原因，文档现在说明:

注意:请注意，三元运算符是一个表达式，它的计算结果不是变量，而是表达式的结果.了解是否要通过引用返回变量很重要.语句返回 $var == 42 ?$a : $b;因此，在按引用返回的函数中将不起作用，并且在更高的 PHP 版本中会发出警告.

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.

另请参阅此错误，其中三元运算符实际上通过引用返回foreach 上下文，当它不应该时.

See also this bug where the ternary operator actually returns by reference in a foreach context, when it shouldn't.

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