内联命名空间变量是否具有内部链接？如果没有，为什么下面的代码工作？ [英] Do inline namespace variables have internal linkage? If not, why does the code below work?

问题描述

此问题与此一直接相关。考虑代码：

This question is directly related to this one. Consider the code:

#include <iostream>

inline namespace N1
{
    int x = 42;
}
int x = 10;

int main()
{
    extern int x;
    std::cout << x; // displays 10
}

显示 10 。如果我删除 extern int x; 声明，那么我们得到一个歧义编译器时间错误

It displays 10. If I remove the extern int x; declaration then we get an ambiguity compiler time error

错误：对x的引用不明确

error: reference to 'x' is ambiguous

问题：为什么代码与 extern int x 声明工作，为什么它停止工作，当我删除它？是因为内联命名空间变量有内部链接？

Question: why does the code work with the extern int x declaration work, and why does it stop working when I remove it? Is it because inline namespace variables have internal linkage?

推荐答案

否。在[basic.link]中没有规定会导致 x 具有内部链接。具体来说，所有其他命名空间具有外部链接，其他是指未命名。也许你在考虑未命名的命名空间？

No. There is no provision in [basic.link] that would cause x to have internal linkage. Specifically, "All other namespaces have external linkage.", and "other" refers to "not unnamed". Perhaps you were thinking of unnamed namespaces?

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