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问题描述

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要从PHP函数中返回引用,必须执行以下操作:

In order to return a reference from a function in PHP one must:

function &func() { return $ref; }
$reference = &func();

我正在尝试从闭包中返回引用。在一个简化的示例中,我要实现的是:

I am trying to return a reference from a closure. In in a simplified example, what I want to achieve is:

$data['something interesting'] = 'Old value';

$lookup_value = function($search_for) use (&$data) {
    return $data[$search_for];
}

$my_value = $lookup_value('something interesting');
$my_value = 'New Value';

assert($data['something interesting'] === 'New Value');

我似乎无法获得从函数工作返回引用的常规语法。

I cannot seem to get the regular syntax for returning references from functions working.

推荐答案

您的代码应如下所示:

$data['something interesting'] = 'Old value';

$lookup_value = function & ($search_for) use (&$data) {
    return $data[$search_for];
};

$my_value = &$lookup_value('something interesting');
$my_value = 'New Value';

assert($data['something interesting'] === 'New Value');

检查退出:

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09-08 06:49