如何调用常量作为函数名称? [英] How to call a constant as a function name?

问题描述

在PHP中，您可以通过在变量内调用函数名称来调用函数.

In PHP, you can call functions by calling their name inside a variable.

function myfunc(){ echo 'works'; }

$func = 'myfunc';
$func(); // Prints "works"

但是，您不能使用常量来做到这一点.

But, you can't do this with constants.

define('func', 'myfunc');

func(); // Error: function "func" not defined

有一些解决方法，例如:

There are workarounds, like these:

$f = func;
$f(); // Prints "works"

call_user_func(func); // Prints "works"

function call($f){ $f(); }
call(func); // Prints "works"

callable 说:

PHP函数的名称以字符串形式传递.可以使用任何内置的或用户定义的功能，语言构造除外.

A PHP function is passed by its name as a string. Any built-in or user-defined function can be used, except language constructs.

似乎没有关于常量值不可调用的信息.

There seems to be nothing about constant values not being callable.

我也尝试检查它，当然，

I also tried to check it, and of course,

var_dump(is_callable(func));

打印 bool(true).

现在，是否有这种解释的解释?据我所知，所有解决方法都依赖于将常量值分配给变量，为什么不能调用常量?

Now, is there an explanation as to why is it this way? As far as I can see all the workarounds rely on assigning the constant value to a variable, by why can't constant be called?

再一次，只是为了使其非常清晰，我不需要调用该函数的方法，甚至在此处提供了一些方法.我想知道为什么 PHP不允许直接通过常量调用函数.

And again, just to make it super clear, I don't need a way to call the function, I even presented some there. I want to know why PHP doesn't allow calling the function directly through the constant.

推荐答案

由于您询问了原因/原因，我猜唯一的答案(可能不会满足您)是:
因为尚未在 https://wiki.php.net/rfc 上提出，讨论和接受，

Since you asked for the why/reason I guess the only answer (which will probably not satisfy you) is:
Because it hasn't been proposed, discussed and accepted on https://wiki.php.net/rfc .

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