c++ - 通过来自另一个翻译单元的指针调用具有内部链接的函数

Question

我们能否在匿名命名空间中声明一个回调函数(从而为其提供内部链接)，知道它将从另一个翻译单元(甚至另一个库)调用？

一些库:

void register_callback(void (*cb)())
{
  ..
  cb();
  ..
}

主程序

namespace {
int foo_cb() { ... } // internal linkage
}

int main()
{
   register_callback(foo_cb);
}

Answer 1

TL;DR:是的，没关系

---

来自 [basic.link](强调我的):

2. 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:

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

3. [...]

4. An unnamed namespace or [...] has internal linkage. [...]. A name having namespace scope that has not been given internal linkage above has the same linkage as the enclosing namespace if it is the name of

   • a function; or

所以基本上链接是名称的属性，而不是对象、函数等的属性。这意味着在未命名的命名空间内声明的函数不能被调用来自另一个翻译单元的名称。通过它的指针调用它没有限制。