c++ - 继承构造函数后查找基类名

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?
};

从直觉上看，这段代码显然是有效的，并且确实可以编译(使用 gcc 和 clang 进行了测试)。

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类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，C 是 derived，而 f 是使用 base 在 base foo() 中。

“声明集”在p3中定义:

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 告诉我们什么进入了声明集:

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 是名称base (f) 在derived (C)。该段落继续举例说明声明 not 满足发生查找的语言构造的要求的含义，但没有任何内容可以排除 using base::来自此查找的基础。

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 指定了哪些成员呢？在我看来，[class.qual] p2 回答了这个问题:

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“不忽略函数名称的查找”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.

这些都不是所讨论的名称查找的情况，所以在我看来这一段适用，并且说 using base::base 指定构造函数(这也是你鉴于它是继承的构造函数声明，我会凭直觉期待)。

在派生类范围内找到声明(指定基类构造函数)后，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类范围内找到结果，因此它不会继续在基类范围内查找(它会在哪里找到 injected-class-name 基地)。 [顺便说一句，即使名称查找继续进入基类范围，我也看不到任何可以消除构造函数和注入(inject)类名称之间歧义的内容]。

我的推理哪里出了问题？

Answer 1

该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.