c++ - const 引用在 C++ 中是否有外部链接？

Question

根据 C++ 1998 标准第 3.5 节的第 3 条，const 引用具有内部链接。

A name having namespace scope (3.3.5) has internal linkage if it is the name of

• an object, reference, function or function template that is explicitly declared static or,

• an object or reference that is explicitly declared const and neither explicitly declared extern nor previously declared to have external linkage; or

• a data member of an anonymous union.

但是为什么编译下面的代码会产生多重定义冲突呢？

// a.cpp
const int& a = 1;

int main()
{
    return 0;
}

// b.cpp
const int& a = 1;

然后编译代码。

$ g++ a.cpp b.cpp
/tmp/ccb5Qi0M.o:(.bss+0x0): multiple definition of `a'
/tmp/ccD9vrzP.o:(.bss+0x0): first defined here
collect2: error: ld returned 1 exit status

如果把const引用改成const，如下

// a.cpp
const int a = 1;

int main()
{
    return 0;
}

// b.cpp
const int a = 1;

编译就OK了。

Answer 1

引用本身不是const，只是它引用的对象；所以(可以说)这条规则没有给出引用内部链接。

将引用声明为 const 是没有意义的。 C++11标准明确了措辞:

a variable that is explicitly declared const or constexpr and neither explicitly declared extern nor previously declared to have external linkage

没有提及声明为 const 的引用的荒谬概念。