继承构造函数后查找基 class 名称

Question

考虑以下代码：

struct base {};

struct derived : public base {
    using base::base;
    base foo() const;  // how does name lookup on 'base' here work?
};

直觉上，很明显这段代码是有效的，并且确实可以编译（用 gcc 和 clang 测试过）。

但是，我想了解标准中的哪些内容使其有效。具体来说，我想了解 base foo() 中 base 的名称查找如何找到基 class 类型而不是继承的构造函数。

这是我对标准措辞的分析，表明它应该解析为构造函数。这可能是错误的，但我想了解我哪里出错了。

我从 [class.member.lookup] p1 开始：

Member name lookup determines the meaning of a name (id-expression) in a class scope. [...] For an id-expression, name lookup begins in the class scope of this

p7 告诉我们名称查找的结果是什么：

The result of name lookup for [a member name] f in [a class scope] C is the declaration set of S(f, C)

我正在尝试遵循此过程，其中 C 是 derived，f 是 base foo() 中 base 的使用。

"Declaration set" 定义在 p3:

The lookup set for f in C, called S(f, C), consists of two component sets: the declaration set, a set of members named f; [...]

p4 告诉我们什么进入声明集：

If C contains a declaration of the name f, the declaration set contains every declaration of f declared in C that satisfies the requirements of the language construct in which the lookup occurs.

using base::base是derived(C)中名字base(f)的声明。该段继续举例说明声明 not 满足发生查找的语言结构的要求的含义，但没有任何内容可以排除 using base::base 从这个查找。

接下来，在 p3 中，我们被告知如何处理声明集中的 using-declarations：

In the declaration set, using-declarations are replaced by the set of designated members that are not hidden or overridden by members of the derived class

那么using base::base指派了哪些成员呢？在我看来 [class.qual] p2:

In a lookup in which function names are not ignored and the nested-name-specifier nominates a class C:

if the name specified after the nested-name-specifier, when looked up in C, is the injected-class-name of C, or

in a using-declaration that is a member-declaration, if the name specified after the nested-name-specifier is the same as identifier [...] in the last component of the nested-name-specifier

the name is instead considered to name the constructor of class C.

有一个脚注阐明了 "lookup in which function names are not ignored" 的含义：

Lookups in which function names are ignored include names appearing in a nested-name-specifier, an elaborated-type-specifier, or a base-specifier.

None 这些是有问题的名称查找的情况，所以在我看来这一段适用，并说 using base::base 指定构造函数（这也是你的d 直观地期待，因为它是继承构造函数声明）。

在派生的 class 范围内找到一个声明（指定基 class 构造函数），我们继续遵循 [class.member.lookup] p4:

If the resulting declaration set is not empty, the subobject set contains C itself, and calculation is complete.

也就是说，由于名称查找在派生的 class 范围内找到结果，因此它不会继续在基础 class 范围内查找（它将在其中找到 注入-class-名称 base)。 [顺便说一句，即使名称查找继续进入基础 class 范围，我也看不到任何可以消除构造函数和 injected-class-name].

我的推理哪里出了问题？

Answer 1

标准煞费苦心地指出构造函数没有名称。无法通过名称查找找到它，因为它没有名称。

C++11 §12.1/1

” Constructors do not have names.

C+11 §12.1/2

” Because constructors do not have names, they are never found during name lookup.

继承构造函数后查找基 class 名称

Lookup of base class name after inheriting constructor

c++

name-lookup

c++11

inheriting-constructors

injected-class-name