SFINAE: checking the existence of a function break

问题:

I want to check for the existence of a function in a specific namespace using SFINAE. I have found SFINAE to test a free function from another namespace which does the job, but there are some things I don't understand.

Currently I have this working code, straight from the linked question:

// switch to 0 to test the other case
#define ENABLE_FOO_BAR 1

namespace foo {
  #if ENABLE_FOO_BAR
    int bar();
  #endif
}

namespace detail_overload {
  template<typename... Args> void bar(Args&&...);
}
namespace detail {
  using namespace detail_overload;
  using namespace foo;
  template<typename T> decltype(bar()) test(T);
  template<typename> void test(...);
}
static constexpr bool has_foo_bar = std::is_same<decltype(detail::test<int>(0)), int>::value;

static_assert(has_foo_bar == ENABLE_FOO_BAR, "something went wrong");

(the ENABLE_FOO_BAR macro is just for testing purpose, in my real code I don't have such a macro available otherwise I wouldn't be using SFINAE)

---

However, as soon as I put detail_overload::bar() in any other namespace (adjusting the using directive as needed), the detection breaks silently and the static_assert kicks in when foo::bar() exists. It only works when the "dummy" bar() overload is directly in the global namespace, or part of the ::detail_overload namespace (note the global :: scope).

// breaks
namespace feature_test {
  namespace detail_overload {
    template<typename... Args> void bar(Args&&...);
  }
  namespace detail {
    using namespace detail_overload;
    using namespace foo;
    //...

// breaks
namespace feature_test {
  template<typename... Args> void bar(Args&&...);
  namespace detail {
    using namespace foo;
    //...

// breaks
namespace detail {
  namespace detail_overload {
    template<typename... Args> void bar(Args&&...);
  }
  using namespace detail_overload;
  using namespace foo;
  //...

// works
template<typename... Args> void bar(Args&&...);
namespace feature_test {
  namespace detail {
    using namespace foo;
    //...

// works
namespace detail_overload {
  template<typename... Args> void bar(Args&&...);
}
namespace feature_test {
  namespace detail {
    using namespace detail_overload;
    using namespace foo;
    //...

I realize this is the very same problem as the question I linked to, and as mentioned I already have a working solution, but what is not addressed there is why precisely does this happen?

As a side question, is there any way to achieve correct SFINAE detection without polluting the global namespace with either bar() or a detail_overload namespace? As you can guess from the non-working examples, I'd like to neatly wrap everything in a single feature_test namespace.

回答1:

I'll change it slightly so the fall-back declaration of bar isn't a template (= shorter code), and don't use SFINAE as this is purely a name lookup issue.

namespace foo {
    int bar(int);
}

namespace feature_test {
    namespace detail_overload {
        void bar(...);
    }

    namespace detail {
        using namespace detail_overload;
        using namespace foo;

        void test() { bar(0); } // (A)
    }
}

In line (A), the compiler needs to find the name bar. How is it looked up? It's not argument-dependent, so it must be unqualified lookup: [basic.lookup.unqual]/2

The declarations from the namespace nominated by a using-directive become visible in a namespace enclosing the using-directive; see 7.3.4. For the purpose of the unqualified name lookup rules described in 3.4.1, the declarations from the namespace nominated by the using-directive are considered members of that enclosing namespace.

Note they become in an enclosing namespace, not the enclosing namespace. The details from [namespace.udir]/2 reveal the issue:

[...] During unqualified name lookup (3.4.1), the names appear as if they were declared in the nearest enclosing namespace which contains both the using-directive and the nominated namespace.

That is, for the name lookup of bar inside test:

namespace foo {
    int bar(int);
}

// as if
using foo::bar;
namespace feature_test {
    namespace detail_overload {
        void bar(...);
    }

    // as if
    using detail_overload::bar;
    namespace detail {
        // resolved
        // using namespace detail_overload;
        // using namespace foo;

        void test() { bar(0); } // (A)
    }
}

Therefore, the name bar found in feature_test hides the name (not) found in the global scope.

Note: Maybe you can hack around this issue with argument-dependent name lookup (and a second SFINAE). If something comes to my mind, I'll add it.

回答2:

In addition to DyP's answer and following his comment:

If your function bar took any arguments, you could make use of dependent name lookup to make it work (w/o a second overload of bar).

Indeed in my real code bar() does take arguments.

As a side question, is there any way to achieve correct SFINAE detection without polluting the global namespace...

So yes, dependent name lookup works like a charm. For the sake of completeness, and in case it can help others in the future, here's my now perfectly working code:

#define ENABLE_FOO_BAR 1

namespace foo {
  #if ENABLE_FOO_BAR
    int bar(int);
  #endif
}

namespace feature_test {
  namespace detail {
    using namespace foo;
    template<typename T> decltype(bar(std::declval<T>())) test(int);
    template<typename> void test(...);
  }
  static constexpr bool has_foo_bar = std::is_same<decltype(detail::test<int>(0)), int>::value;
  static_assert(has_foo_bar == ENABLE_FOO_BAR, "something went wrong");
}

All credit goes to DyP, I don't believe I'd have thought about this by myself.