为什么这个 ADL 案例有效？

Question

find_type怎么知道函数typemap在哪里？
它收到的参数不是来自那个命名空间，而是来自 std 命名空间！

#include <type_traits>
#include <memory>

namespace lib {
    template<typename T>
    struct find_type {
        using type = decltype(typemap(std::declval<T>()));
    };
}

namespace test {
    struct Test {};
    auto typemap(std::unique_ptr<Test>) -> int;    
}

static_assert(std::is_same<int, lib::find_type<std::unique_ptr<test::Test>>::type>::value, "");

这段代码如何工作？允许这样做的规则是什么？

我用 GCC 6.3 和 clang 3.9.1 测试了它。

Answer 1

在 C++ 标准 N4618 §3.4.2 [basic.lookup.argdep] (2.2)

If T is a class type (including unions), its associated classes are: the class itself; the class of which it is a member, if any; and its direct and indirect base classes. Its associated namespaces are the innermost enclosing namespaces of its associated classes. Furthermore, if T is a class template specialization, its associated namespaces and classes also include: the namespaces and classes associated with the types of the template arguments provided for template type parameters (excluding template template parameters); the namespaces of which any template template arguments are members; and the classes of which any member templates used as template template arguments are members.

typemap 的参数是 std::unique_ptr<test::Test>，因此命名空间 test 被考虑用于名称查找。

Answer 2

The argument it receives is not from that namespace, it's from the std namespace!

不是全部！

using type = decltype(typemap(std::declval<T>()));

这是：

using type = decltype(typemap(std::declval<std::unique_ptr<test::Test>>()));

里面有一个test::，所以命名空间test也被搜索了。

为什么这个 ADL 案例有效？

Why this ADL case is working?

c++

language-lawyer

argument-dependent-lookup

c++11