c++ - 静态常量成员初始化中的sizeof

Question

我有这样的代码:

class A
{
public:

    unsigned long a;
    static const unsigned long b = sizeof(a); // "error C2327: 'A::a' : is not a type name, static, or enumerator" in VC++
};

我在 VC++ 中遇到编译器错误，而在 IAR 中没有错误。哪个编译器是正确的，C++ 标准是怎么说的？

Answer 1

您的 MSVS 版本很旧，因此基于此，并假设它们默认为 C++03，它们拒绝您的代码是正确的。我会引用 n1905 ，就我们的目的而言，它非常接近 C++03 标准。

9.4 [class.static] (emphasis mine)

If an unqualified-id (5.1) is used in the definition of a static member following the member’s declarator-id, and name lookup (3.4.1) finds that the unqualified-id refers to a static member, enumerator, or nested type of the member’s class (or of a base class of the member’s class), the unqualified-id is transformed into a qualified-id expression in which the nested-name-specifier names the class scope from which the member is referenced. The definition of a static member shall not use directly the names of the non-static members of its class or of a base class of its class (including as operands of the sizeof operator). The definition of a static member may only refer to these members to form pointer to members (5.3.1) or with the class member access syntax (5.2.5).