PHP匿名函数变量作为参考 [英] PHP anonymous function variable as reference

问题描述

在使用Laravel框架(更具体地-表单宏)时，我偶然发现了一个奇怪的错误.

While working with Laravel framework, more specific - Form macros, I stumbled upon a weird error.

起初，我认为Laravel出了点问题，但是后来我把所有内容都从上下文中移走了:

At first, I thought it's something wrong with Laravel, but then I took everything out of context:

<?php

// placeholder function that takes variable as reference
$function = function(&$reference)
{
    // append to variable
    $reference = $reference . ':' . __METHOD__;
};

// test with straight call
$variable = 'something';
$function($variable);
echo $variable;


// test with call_user_func(), that gets called in Laravels case
$variable = 'something'; // reset
call_user_func($function, $variable);
echo $variable;

第一次正确调用$function时，第二次尝试产生(摘录自键盘):

While the first call to $function executes properly, the second try with call_user_func(), produces (excerpt from Codepad):

Warning: Parameter 1 to {closure}() expected to be a reference, value given
PHP Warning: Parameter 1 to {closure}() expected to be a reference, value given

提琴:键盘@ Viper-7

在撰写本文时，我想到了call_user_func_array():在这里摆弄，但存在相同的错误产生了.

While writing this, I thought about call_user_func_array(): fiddle here, but the same error is produced.

我对引用有什么误解吗，或者这是PHP的错误?

Have I got something wrong about references or is this a bug with PHP?

推荐答案

尽管从技术上讲，它是call_user_func的错误，但我将其称为PHP的错误.该文档确实提到了这一点，但可能不是很启发性的:

I would call this a bug with PHP, although it's technically a bug with call_user_func. The documentation does mention this, but perhaps not in a very enlightening way:

请注意，call_user_func()的参数不会传递 参考.

Note that the parameters for call_user_func() are not passed by reference.

说 call_user_func()的参数不通过引用传递可能更清楚(但请注意，从技术上讲，根本不需要说什么；该信息也嵌入在功能签名).

It would be perhaps clearer to say that the arguments to call_user_func() are not passed by reference (but note that technically it's not necessary to say anything at all; this information is also embedded in the function signature).

无论如何，这意味着，当call_user_func最终调用其目标可调用对象时，所传递的参数的ZVAL(所有类型的值的PHP引擎内部数据结构)都不会标记为"-a参考"；闭包在运行时会对此进行检查并抱怨，因为其签名表明该参数必须是引用.

In any case, this means is that when call_user_func finally gets to invoking its target callable, the ZVAL (PHP engine internal data structure for all types of values) for the argument being passed is not marked as "being-a-reference"; the closure checks this at runtime and complains because its signature says that the argument must be a reference.

在PHP中< 5.4.0可以通过使用按引用传递的呼叫时间来解决此问题:

In PHP < 5.4.0 it is possible to work around this by using call-time pass by reference:

 call_user_func($function, &$variable);

但这会产生E_DEPRECATED警告，因为按引用的调用时传递是一项已弃用的功能，并且由于完全删除了该功能，将导致PHP 5.4中的致命错误.

but this produces an E_DEPRECATED warning because call-time pass by reference is a deprecated feature, and will flat out cause a fatal error in PHP 5.4 because the feature has been removed completely.

结论:没有以这种方式使用call_user_func的好方法.

Conclusion: there is no good way to use call_user_func in this manner.

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