C ++名称查找-来自标准的示例 [英] C++ name lookup - example from the standard

问题描述

我需要从标准[basic.lookup.unqual]/3中对该示例进行解释:

I need an explanation on this example from standard [basic.lookup.unqual]/3:

typedef int f;
namespace N {
struct A {
    friend void f(A &);
    operator int();
    void g(A a) {
        int i = f(a); // f is the typedef, not the friend
                      // function: equivalent to int(a)
    }
};
}

我认为void N::f(A &)比int(a)更近，因为它不涉及类型转换运算符.但是我不确定，因为该标准仅包含类型转换"的一个实例.

I would argue that void N::f(A &) is closer than int(a) because it does not involve the type-cast operator. I cannot however be sure because the standard contains only one instance of "type cast".

顺便说一句，在MSVC2015中编译该代码失败(但它可以在clang和g ++中工作).

By the way, compiling that code fails in MSVC2015 (but it works in clang and g++).

错误C2440'正在初始化':无法从'void'转换为'int'

Error C2440 'initializing': cannot convert from 'void' to 'int'

---

更新 解决一些评论.

---

UPDATE Addressing some of the comments.

1. 类型转换在形式上被称为类型转换"，在(12.3)中进行了介绍(感谢jefffrey).

1. Type casting is formally known as "type conversions" and they are covered in (12.3) (thanks jefffrey).

我正在寻找的是语法解析的描述.特别是为什么postfix-expression ( expression-list opt )被simple-type-specifier ( expression-list opt )践踏.由于根据(5.2)，这两个表达式都是从左到右求值的.因此，在计算::N::A::g中的表达式时，在(a)之前的两个候选中，::N::f应该比::f更近.

What I am looking for is a description of the syntax parsing. In particular, why postfix-expression ( expression-list opt ) is trampled over by simple-type-specifier ( expression-list opt ). Since according to (5.2), both of those expression are evaluated left-to-right. Hence, out of the two candidates to be before the (a), ::N::f should be closer than ::f when evaluating expressions in ::N::A::g.

推荐答案

这里发生了一些事情.如包含示例的注释所述(3.4/3):

There are a few things going on here. As the note containing the example says (3.4/3):

出于确定(解析期间)表达式是否为函数调用的后缀表达式的目的，通常使用名称查找规则. 3.4.2中的规则对表达式的句法解释没有影响.

For purposes of determining (during parsing) whether an expression is a postfix-expression for a function call, the usual name lookup rules apply. The rules in 3.4.2 have no effect on the syntactic interpretation of an expression.

因此，首先我们需要使用 don'的名称查找规则来了解f是简单类型说明符还是后缀表达式 t 包括3.4.2节.在这些规则下，功能N::f不可见.

So first we need to know whether f is a simple-type-specifier or a postfix-expression, using name lookup rules that don't include section 3.4.2. And under these rules, the function N::f is not visible.

7.3.1.2/3:

7.3.1.2/3:

如果非本地类中的friend声明首先声明了一个类，函数，类模板或函数模板，则该朋友是最内部封闭的命名空间的成员. friend声明本身不会使名称对不合格的查找(3.4.1)或合格的查找(3.4.3)可见.

If a friend declaration in a non-local class first declares a class, function, class template or function template the friend is a member of the innermost enclosing namespace. The friend declaration does not by itself make the name visible to unqualified lookup (3.4.1) or qualified lookup (3.4.3).

因此，不合格的查找根本看不到任何N::f声明，而仅找到::f，它是类型名.因此语法是简单类型说明符 ( 表达式列表 _opt )，而不是后缀表达式 em> ( 表达式列表 _opt )，并且不依赖于参数的查找(3.4.2).

Therefore the unqualified lookup does not see any declaration of N::f at all, and finds only ::f, which is a typename. So the syntax is simple-type-specifier ( expression-list _opt ) and not postfix-expression ( expression-list _opt ), and argument-dependent lookup (3.4.2) does not apply.

((如果不合格的查找找到了一个函数名，则3.4.2将适用，并且尽管没有声明，也可以将N::f包括在候选列表中.如果N::f以前有一个声明不是friend声明，它将赢得不合格的查找.)

( If the unqualified lookup had found a function name, 3.4.2 would apply and would be able to include N::f in the candidate list despite the lack of declaration. If N::f had a previous declaration other than the friend declaration, it would have won the unqualified lookup. )

这篇关于C ++名称查找-来自标准的示例的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！