c++ - 为什么在使用 `using directive` 时 ADL 不工作？

Question

Here is a similar question , 但在这个问题上它有效，但是在以下情况下它失败了，为什么？

namespace A
{
  int k;
}
namespace B
{
  class test{};
  void k(const test&){/*do something*/}
}

int main()
{
  using namespace A;
  k(B::test());//compile error
}  

错误信息是:“‘A::k’不能用作函数”(gcc 6.3.0)

也就是说，编译器不会尝试执行ADL，也永远不会在namespace B 中找到void k(const test&)

不过，我认为ADL应该在这种情况下工作，因为上面的代码不属于以下情况:

引自cppref

First, the argument-dependent lookup is not considered if the lookup set produced by usual unqualified lookup contains any of the following:
1) a declaration of a class member
2) a declaration of a function at block scope (that's not a using-declaration)
3) any declaration that is not a function or a function template (e.g. a function object or another variable whose name conflicts with the name of the function that's being looked up)

更准确地说，这里的using namespace A 没有引入任何声明:
引自 cppref

Using-directive does not add any names to the declarative region in which it appears (unlike the using-declaration), and thus does not prevent identical names from being declared.

Answer 1

函数调用的名称查找有两部分:

• 正常的不合格查找
• 日常事件能力

根据 N4659 [basic.lookup.argdep]/3，正常的非限定查找首先发生；如果找到正常的不合格查找，则 ADL 阶段不会继续:

• a declaration of a class member, or
• a block-scope function declaration that is not a using-declaration, or
• a declaration that is neither a function nor a function template.

在您的代码中，正常的非限定查找会找到 A::k，如 your previous question 中所述。 .因此此代码不会发生 ADL。