c++ - 名称查找后跟在合格的 declarator-id 之后

Question

我试图理解以下引用的含义 (3.4.3/3 N3797):

names following the qualified-id are looked up in the scope of the member’s class or namespace.

namespace A
{
    class C
    {
    public:
        static const int a=7;
        static int b;
    };
}

int A::C::b=a; //7

static int b; 的范围仅由b 的声明点之后的声明区域组成。实际上:

The potential scope of a name declared in a class consists not only of the declarative region following the name’s point of declaration, but also of all function bodies, default arguments, exception-specifications, and brace-or-equal-initializers of non-static data members in that class

这意味着static const int a=7;不属于static int b;的范围。因此 static const int a=7 无法在 int A::C::b=a; 中找到。

这是标准中的错字还是我的误解？

Answer 1

This implies that static const int a=7; does not belong to the scope of static int b;. Hence the static const int a=7 cannot be found in the int A::C::b=a;.

没有。它准确地暗示了您在那里可以读到的内容:在类中声明的名称的潜在范围还包含非静态数据成员的函数体等。这与上面的引用并不冲突 - 静态数据成员的声明区域(和范围)仍然包含它本身声明的类的范围。

你自己引用了相关部分:

names following the qualified-id are looked up in the scope of the member’s class or namespace

因此，因为在这个代码片段中

int A::C::b=a;

a用在declarator-id之后，在类中查找找到。