c++ - 类成员转换函数-id

Question

我正在寻找 Stadard 中描述以下行为的引述:

以下规则用于 conversion-type-id 查找 (3.4.6/7):

If the id-expression is a conversion-function-id, its conversion-type-id is first looked up in the class of the object expression and the name, if found, is used. Otherwise it is looked up in the context of the entire postfix-expression.

考虑以下示例:

#include <iostream>

class J{ public: static const char a = 'j'; };

typedef J Y;

class C
{
public:
    operator Y(){ std::cout << Y::a; }
};

int main()
{ 
    typedef Y Z;
    C *c= new C(); 
    c -> operator Z(); //C::operator Y is invoked
}

我不明白。上面的引述并没有描述这种行为。它仅描述了对 conversion-type-id 的查找，但未描述 conversion-function-id 本身。

conversion-function-id 的查找规则是什么？

Answer 1

如果显式使用conversion-function-id，如示例表达式c->operator Z()，查找规则与任何其他类(class)成员。一个技巧是第 3 节第 8 段的定义:

Two names are the same if

• they are identifiers composed of the same character sequence, or

• they are operator-function-ids formed with the same operator, or

• they are conversion-function-ids formed with the same type, or

• they are template-ids that refer to the same class or function, or

• they are the names of literal operators formed with the same literal suffix identifier.

此处的第三个选项是表达式中使用的名称 operator Z 找到类成员 C::operator Y() 的原因。它是访问类中的一个成员，与表达式的 conversion-function-id 同名。