c++ - ODR和内部链接

Question

假设我在一个程序中有两个编译单元，每个编译单元都声明了一个非内联函数，具有相同的签名，但实现不同，例如

// a.cpp
namespace internal {
    int foo(int a) {
        return a+1;
    }
}

int main() {
}

和

// b.cpp
namespace internal {
    int foo(int b) {
        return b+2;
    }
}

编译/链接这个(g++ 4.8.3 with -std=c++11)，我得到一个错误

b.cpp:(.text+0x0): multiple definition of `internal::foo(int)'

这是意料之中的，因为据我所知，这只是违反了 one definition rule :

One and only one definition of every non-inline function or variable that is odr-used (see below) is required to appear in the entire program (including any standard and user-defined libraries).

现在，将 namespace internal 更改为未命名的命名空间，错误消失了。直觉上，这对我来说很有意义，因为我正在更改 external to internal linkage 的函数:

Any of the following names declared at namespace scope have external linkage unless the namespace is unnamed or is contained within an unnamed namespace (since C++11): variables and functions not listed above (that is, functions not declared static [...]) ...

和

[A]ll names declared in unnamed namespace or a namespace within an unnamed namespace, even ones explicitly declared extern, have internal linkage.

但是，我无法在一个定义规则中找到任何免除具有内部链接的函数的内容。因此，我的问题是:我的直觉推理是否正确，或者我是否仍然违反具有内部链接的函数的单一定义规则(并且编译器/链接器不再报告它)？此外，标准(或 cppreference.com :))在哪里说明它是否可以？

Answer 1

n4713

§6.5 Program and linkage [basic.link]

A name is said to have linkage when it might denote the same object, reference, function, type, template, namespace or value as a name introduced by a declaration in another scope:

(2.1) — When a name has external linkage, the entity it denotes can be referred to by names from scopes of other translation units or from other scopes of the same translation unit.

(2.2) — When a name has internal linkage, the entity it denotes can be referred to by names from other scopes in the same translation unit.

— When a name has no linkage, the entity it denotes cannot be referred to by names from other scopes.

这基本上是说(在未命名命名空间的情况下)来自 a.cpp 的名称 foo 和来自 的名称 foo >b.cpp 分别引用不同的实体。所以您没有同一对象的两个定义，因此不违反 ODR。