问题描述
为什么不可能引用void？我在C ++标准中发现的唯一内容是以下行，位于 8.3.2.1
Why is it impossible to have a reference to void? The only thing I found in the C++ Standard is this line, at 8.3.2.1
为什么这样呢？为什么我不能编写一个接受 void& 的通用函数？
Why is it that way? Why can't I write a "generic" function that accept a void&?
只是要清楚一点，我没有想到有用的应用程序，在这种情况下，使用void引用比使用模板要好，但是我只是对此感到好奇
Just to be clear, I have no useful application in mind where using a reference-to-void could be better than using templates, but I'm just curious about the rationale for forbidding this construct.
为了澄清一点，我理解使用空引用是这样与取消引用空指针一样没有意义。但是，我可以将其转换为对 sometype 的引用才能使用，不是吗？实际上，我不明白为什么以下代码段可以工作...
To clarify a little, I understand that using a reference-to-void "as is" would be as meaningless as dereferencing a pointer-to-void. However, I could cast it to a reference-to-sometype in order to use it, couldn't I? In fact, I don't see why the following snippet can work...
void foo(void *data)
{
int *i = reinterpret_cast<int*>(data);
// do something with i
}
...而这一个不能：
...while this one cannot:
void foo(void &data)
{
int &i = reinterpret_cast<int&>(data);
// do something with i
}
推荐答案
如果您确实引用过void，那么您将如何处理它？它不是数字，字符，指针或类似的东西。您假设的泛型函数除了对其地址（而不是其大小）进行任何操作都不能对其执行任何操作。
If you did have a reference to void, what would you do with it? It wouldn't be a number, or a character, or a pointer, or anything like that. Your hypothetical generic function couldn't perform any operation on it, except taking its address (and not its size).
void有两个用途：放弃任何知识类型（如void *），并且不指定任何内容（与void函数返回相反）。在这两种情况下，都不可能说出一个关于虚空的东西，只是它可能有一个地址。
"void" has two uses: to disclaim any knowledge of type (as in void *), and to specify nothing as opposed to something (void function return). In neither case is it possible to say anything about a void something except that it may have an address.
如果您想不出一种有用的方法，而且我不能，那至少是证明某事无用的证据，而这很可能至少是这里理由的一部分。
If you can't think of a way something can be useful, and I can't, that is at least evidence that something is useless, and that may well be at least part of the rationale here.
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