c++ - 非唯一 namespace 限定查找的代码示例

Question

S(X,m) 的定义如下。 3.4.3.2/2:

For a namespace X and name m, the namespace-qualified lookup set S(X, m) is defined as follows: Let S (X, m) be the set of all declarations of m in X and the inline namespace set of X (7.3.1). If S (X, m) is not empty, S(X, m) is S (X, m); otherwise, S(X, m) is the union of S(Ni , m) for all namespaces Ni nominated by using-directives in X and its inline namespace set.

引自 3.4.3.2/3:

Given X::m (where X is a user-declared namespace), or given ::m (where X is the global namespace), if S(X, m) is the empty set, the program is ill-formed. Otherwise, if S(X, m) has exactly one member, or if the context of the reference is a using-declaration (7.3.3), S(X, m) is the required set of declarations of m. Otherwise if the use of m is not one that allows a unique declaration to be chosen from S(X, m), the program is ill-formed.

你能举个例子来证明这个规则吗:

if the use of m is not one that allows a unique declaration to be chosen from S(X, m), the program is ill-formed.

Answer 1

标准草案实际上提供了一个示例，虽然它不是很容易找到，但在第 3 段下，如下所示，给出以下内容:

namespace A {
  using namespace Y;
  void f(int);
  void g(int);
  int i;
}
namespace B {
  using namespace Z;
  void f(char);
  int i;
}
namespace AB {
  using namespace A;
  using namespace B;
  void g();
}

隐藏在示例中的是以下内容:

AB::i++; // i is not declared directly in AB so the rules are
         // applied recursively to A and B,
         // S is { A::i , B::i } so the use is ambiguous
         // and the program is ill-formed