为什么内联未命名命名空间？ [英] Wherefore inline unnamed namespaces?

问题描述

一个快速的专家：C ++ 11允许未命名的命名空间声明为 inline 。这对我来说似乎是多余的;

所以我的问题是这是什么意思：

  inline namespace / * anonymous * / {
 // stuff 
} 
  

与传统

有什么不同？

  namespace / * anonymous * / {
 // stuff 
} 
  

我们知道和爱从C ++ 98？在使用 inline 时，任何人都可以举例说明不同的行为吗？

EDIT：只是为了澄清，因为这个问题已被标记为重复：我不是问一般命名内联命名空间。我理解这里的用例，我认为他们是伟大的。我特别要求声明一个未命名的命名空间为 inline 的含义。因为未命名的命名空间必须总是在一个TU的本地，符号版本理性似乎不适用，所以我很好奇什么添加 inline 实际上 / em>。

---

另外，关于未命名命名空间的标准[7.3.1.1] p>

inline 出现，如果它出现在 unnamed-namespace-

但这似乎对我的非语言律师的眼睛是一个重言式 - 它出现在定义iff它出现在定义！

编辑： Cubbi声称在评论中的奖励积分：

/ p>

标准是说未命名命名空间定义的行为好像被 X 其中 inline 出现在 X 中，如果它出现在 unnamed-namespace-definition

解决方案

这里是我发现的一个用法：

  namespace widgets {inline namespace {
 
 void foo（）; 
 
}} //命名空间
 
 void widgets :: foo（）
 {
} 
  

在这个例子中， foo 有内部链接，我们可以稍后使用 namespace :: function 语法，以确保函数的签名正确。如果你不使用 widgets 命名空间，那么 void foo（）定义将定义一个完全不同的函数。

如果还有另一个名为 foo 在命名空间已经然后这将给你一个歧义，而不是一个讨厌的ODR违反。

A quick one for the gurus: C++11 allows unnamed namespaces to be declared inline. This seems redundant to me; things declared in an unnamed namespace are already used as if they were declared in the enclosing namespace.

So my question is this: what does it mean to say

inline namespace /*anonymous*/ {
    // stuff
}

and how is it different from the traditional

namespace /*anonymous*/ {
    // stuff
}

that we know and love from C++98? Can anyone give an example of different behaviour when inline is used?

EDIT: Just to clarify, since this question has been marked as a duplicate: I'm not asking about named inline namespaces in general. I understand the use-case there, and I think they're great. I'm specifically asking what it means to declare an unnamed namespace as inline. Since unnamed namespaces are necessarily always local to a TU, the symbol versioning rational doesn't seem to apply, so I'm curious about what adding inline actually does.

---

As an aside, the standard [7.3.1.1], regarding unnamed namespaces, says:

inline appears if and only if it appears in the unnamed-namespace-definition

but this seems like a tautology to my non-language lawyer eyes -- "it appears in the definition iff it appears in the definition"! For bonus points, can anyone explain what this bit of standardese is actually saying?

EDIT: Cubbi claimed the bonus point in the comments:

the standard is saying that unnamed-namespace-definition behaves as if it were replaced by X where inline appears in X iff it appears in the unnamed-namespace-definition

解决方案

Here is one use that I have found:

namespace widgets { inline namespace {

void foo();

} } // namespaces

void widgets::foo()
{
}

In this example, foo has internal linkage and we can define the function later on by using the namespace::function syntax to ensure that the function's signature is correct. If you were to not use the widgets namespace then the void foo() definition would define a totally different function. You also don't need to re-open the namespace saving you a level of indentation.

If there is another function called foo in the widgets namespace already then this will give you an ambiguity instead rather than a nasty ODR violation.

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